2. Determination of the coefficients of the series by the Fourier formulas.
Let a periodic function ƒ(x) with a period of 2π be such that it is represented by a trigonometric series converging to a given function in the interval (-π, π), i.e., is the sum of this series:
Suppose that the integral of the function on the left side of this equality is equal to the sum of the integrals of the terms of this series. This will be true if we assume that the number series composed of the coefficients of the given trigonometric series converges absolutely, i.e., the positive number series converges
Series (1) is majorized and can be integrated term by term in the interval (-π, π). We integrate both parts of equality (2):
We calculate separately each integral occurring on the right side:
,
,
In this way, , where
. (4)
Estimation of the Fourier coefficients. (Bugrov)
Theorem 1. Let a function ƒ(x) of period 2π have a continuous derivative ƒ (s) (x) of order s that satisfies the inequality on the entire real axis:
│ ƒ (s) (x)│≤ M s ; (5)
then the Fourier coefficients of the function ƒ satisfy the inequality
Proof. Integrating by parts and taking into account that
ƒ(-π) = ƒ(π), we have
Integrating the right side of (7) sequentially, taking into account that the derivatives ƒ ΄ , …, ƒ (s-1) are continuous and take the same values at the points t = -π and t = π, as well as estimate (5), we obtain the first estimate ( 6).
The second estimate (6) is obtained in a similar way.
Theorem 2. The Fourier coefficients ƒ(x) satisfy the inequality
(8)
Proof. We have
(9)
Introducing a change of variable in this case and taking into account that ƒ(x) is a periodic function, we obtain
Adding (9) and (10), we get
We carry out the proof for b k in a similar way.
Consequence. If the function ƒ(x) is continuous, then its Fourier coefficients tend to zero: a k → 0, b k → 0, k → ∞.
Space of functions with scalar product.
A function ƒ(x) is called piecewise continuous on a segment if it is continuous on this segment, except perhaps for a finite number of points where it has discontinuities of the first kind. Such points can be added and multiplied by real numbers and, as a result, again piecewise-continuous functions on a segment can be obtained.
The scalar product of two piecewise continuous on (a< b) функций ƒ и φ будем называть интеграл
(11)
Obviously, for any piecewise-continuous functions ƒ , φ , ψ the following properties hold:
1) (ƒ , φ) =(φ, ƒ);
2) (ƒ , ƒ) and the equality (ƒ , ƒ) = 0 implies that ƒ(x) =0 on , excluding, perhaps, a finite number of points x;
3) (α ƒ + β φ , ψ) = α (ƒ , ψ) + β (φ , ψ),
where α, β are arbitrary real numbers.
The set of all piecewise continuous functions defined on the interval , for which the scalar product is introduced according to the formula (11), we will denote, and call space
Remark 1.
In mathematics, a space = (a, b) is a set of functions ƒ(x) that are integrable in the Lebesgue sense on together with their squares, for which the scalar product is introduced by formula (11). The space in question is a part of . Space has many of the properties of space, but not all.
Properties 1), 2), 3) imply the important Bunyakovskii inequality | (ƒ , φ) | ≤ (ƒ , ƒ) ½ (φ , φ) ½ , which in the language of integrals looks like this:
Value
is called the norm of the function f.
The norm has the following properties:
1) || f || ≥ 0, while the equality can only be for the zero function f = 0, i.e., the function equal to zero, except, perhaps, for a finite number of points;
2) || ƒ + φ || ≤ || ƒ(x) || || φ ||;
3) || α ƒ || = | α | · || ƒ ||,
where α is a real number.
The second property in the language of integrals looks like this:
and is called the Minkowski inequality.
It is said that a sequence of functions ( f n ), belongs to , converges to a function belongs in the sense of the mean square on (or else in the norm ), if
Note that if the sequence of functions ƒ n (x) converges uniformly to the function ƒ(x) on the segment , then for sufficiently large n the difference ƒ(x) - ƒ n (x) in absolute value must be small for all x from the segment .
If ƒ n (x) tends to ƒ(x) in the mean square sense on the segment , then the indicated difference may not be small for large n everywhere on . In some places of the segment, this difference can be large, but it is only important that the integral of its square over the segment be small for large n.
Example. Let on a given continuous piecewise linear function ƒ n (x) (n = 1, 2,…) shown in the figure, and
(Bugrov, p. 281, fig. 120)
For any natural n
and, consequently, this sequence of functions, although it converges to zero as n → ∞, is not uniform. Meanwhile
i.e., the sequence of functions (f n (x)) tends to zero in the sense of the mean square on .
From the elements of some sequence of functions ƒ 1 , ƒ 2 , ƒ 3 ,… (belonging to ) we construct a series
ƒ 1 + ƒ 2 + ƒ 3 +… (12)
The sum of its first n members
σ n = ƒ 1 + ƒ 2 + … + ƒ n
there is a function that belongs to . If it happens that in there exists a function ƒ such that
|| ƒ-σ n || → 0 (n → ∞),
then we say that series (12) converges to the function ƒ in the mean square sense and write
ƒ = ƒ 1 + ƒ 2 + ƒ 3 +…
Remark 2.
One can consider the space = (a, b) of complex-valued functions ƒ(x) = ƒ 1 (x) + iƒ 2 (x), where ƒ 1 (x) and ƒ 2 (x) are real piecewise continuous functions. In this space, functions are multiplied by complex numbers and the scalar product of functions ƒ(x) = ƒ 1 (x) + iƒ 2 (x) and φ(x) = φ 1 (x) + i φ 2 (x) is defined as follows:
and the norm ƒ is defined as the value
Fourier series is a representation of an arbitrarily taken function with a specific period as a series. In general terms, this solution is called the decomposition of an element in an orthogonal basis. The expansion of functions in a Fourier series is a fairly powerful tool for solving various problems due to the properties of this transformation when integrating, differentiating, as well as shifting an expression in an argument and convolution.
A person who is not familiar with higher mathematics, as well as with the works of the French scientist Fourier, most likely will not understand what these “series” are and what they are for. Meanwhile, this transformation has become quite dense in our lives. It is used not only by mathematicians, but also by physicists, chemists, physicians, astronomers, seismologists, oceanographers and many others. Let us also take a closer look at the works of the great French scientist, who made a discovery ahead of his time.
Man and the Fourier Transform
>Fourier series is one of the methods (along with analysis and others) This process occurs every time a person hears any sound. Our ear automatically transforms elementary particles in an elastic medium, they are decomposed into rows (along the spectrum) of successive values of the volume level for tones of different heights. Next, the brain turns this data into sounds familiar to us. All this happens in addition to our desire or consciousness, by itself, but in order to understand these processes, it will take several years to study higher mathematics.
More on the Fourier Transform
>The Fourier transform can be carried out by analytical, numerical and other methods. Fourier series refer to the numeral way of decomposing any oscillatory processes - from ocean tides and light waves to cycles of solar (and other astronomical objects) activity. Using these mathematical techniques, it is possible to analyze functions, representing any oscillatory processes as a series of sinusoidal components that go from minimum to maximum and vice versa. The Fourier transform is a function that describes the phase and amplitude of sinusoids corresponding to a specific frequency. This process can be used to solve very complex equations that describe dynamic processes that occur under the influence of thermal, light or electrical energy. Also, Fourier series make it possible to isolate the constant components in complex oscillatory signals, which made it possible to correctly interpret the obtained experimental observations in medicine, chemistry and astronomy.
History reference
>The founding father of this theory is the French mathematician Jean Baptiste Joseph Fourier. This transformation was subsequently named after him. Initially, the scientist applied his method to study and explain the mechanisms of heat conduction - the spread of heat in solids. Fourier suggested that the original irregular distribution can be decomposed into the simplest sinusoids, each of which will have its own temperature minimum and maximum, as well as its own phase. In this case, each such component will be measured from minimum to maximum and vice versa. The mathematical function that describes the upper and lower peaks of the curve, as well as the phase of each of the harmonics, is called the Fourier transform of the temperature distribution expression. The author of the theory reduced the general distribution function, which is difficult to describe mathematically, to a very convenient series of cosine and sine, which sum up to give the original distribution.
The principle of transformation and the views of contemporaries
>The scientist's contemporaries - the leading mathematicians of the early nineteenth century - did not accept this theory. The main objection was Fourier's assertion that a discontinuous function describing a straight line or a discontinuous curve can be represented as a sum of sinusoidal expressions that are continuous. As an example, consider Heaviside's "step": its value is zero to the left of the gap and one to the right. This function describes the dependence of the electric current on the time variable when the circuit is closed. The contemporaries of the theory at that time had never encountered such a situation, when a discontinuous expression would be described by a combination of continuous, ordinary functions, such as an exponential, sinusoid, linear or quadratic.
What confused French mathematicians in Fourier theory?
>After all, if the mathematician was right in his statements, then by summing the infinite trigonometric Fourier series, one can obtain an exact representation of the stepwise expression even if it has many similar steps. At the beginning of the nineteenth century, such a statement seemed absurd. But despite all the doubts, many mathematicians have expanded the scope of the study of this phenomenon, taking it beyond the scope of studies of thermal conductivity. However, most scientists continued to be tormented by the question: "Can the sum of the sinusoidal series converge to the exact value of the discontinuous function?"
Fourier Series Convergence: An Example
>The question of convergence is raised whenever it is necessary to sum infinite series of numbers. To understand this phenomenon, consider a classic example. Can you ever reach the wall if each successive step is half the size of the previous one? Suppose you are two meters from the goal, the first step brings you closer to the halfway point, the next one to the three-quarters mark, and after the fifth step you will cover almost 97 percent of the way. However, no matter how many steps you take, you will not achieve the intended goal in a strict mathematical sense. Using numerical calculations, it can be shown that in the end it is possible to approach an arbitrarily small given distance. This proof is equivalent to demonstrating that the total value of one-half, one-fourth, etc. will tend to one.
A Question of Convergence: The Second Coming, or Lord Kelvin's Appliance
>This question was raised again at the end of the nineteenth century, when Fourier series were tried to be used to predict the intensity of ebb and flow. At this time, Lord Kelvin invented a device, which is an analog computing device that allowed sailors of the military and merchant fleet to track this natural phenomenon. This mechanism determined the sets of phases and amplitudes from a table of tide heights and their corresponding time moments, carefully measured in a given harbor during the year. Each parameter was a sinusoidal component of the tide height expression and was one of the regular components. The results of the measurements were entered into Lord Kelvin's calculator, which synthesized a curve that predicted the height of the water as a function of time for the next year. Very soon similar curves were drawn up for all the harbors of the world.
And if the process is broken by a discontinuous function?
>At that time, it seemed obvious that a tidal wave predictor with a large number of counting elements could calculate a large number of phases and amplitudes and thus provide more accurate predictions. Nevertheless, it turned out that this regularity is not observed in those cases when the tidal expression to be synthesized contained a sharp jump, that is, it was discontinuous. In the event that data is entered into the device from the table of time moments, then it calculates several Fourier coefficients. The original function is restored thanks to the sinusoidal components (according to the found coefficients). The discrepancy between the original and restored expression can be measured at any point. When carrying out repeated calculations and comparisons, it can be seen that the value of the largest error does not decrease. However, they are localized in the region corresponding to the discontinuity point, and tend to zero at any other point. In 1899, this result was theoretically confirmed by Joshua Willard Gibbs of 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 in a certain interval. In general, Fourier series, if the original function is the result of a real physical measurement, always converge. Questions of the convergence of this process for specific classes of functions have led to the emergence of new sections in mathematics, for example, the theory of generalized functions. It is associated with such names as L. Schwartz, J. Mikusinsky and J. Temple. Within the framework of this theory, a clear and precise theoretical basis was created for such expressions as the Dirac delta function (it describes an area of a single area concentrated in an infinitely small neighborhood of a point) and the Heaviside “step”. Thanks to this work, Fourier series became applicable to solving equations and problems in which intuitive concepts appear: a point charge, a point mass, magnetic dipoles, and also a concentrated load on a beam.
Fourier method
>Fourier series, in accordance with the principles of interference, begin with the decomposition of complex forms into simpler ones. For example, a change in heat flow is explained by its passage through various obstacles made of irregularly shaped heat-insulating material or a change in the surface of the earth - an earthquake, a change in the orbit of a celestial body - the influence of planets. As a rule, similar equations describing simple classical systems are elementarily solved for each individual wave. Fourier showed that simple solutions can also be summed to give solutions to more complex problems. Expressed in the language of mathematics, Fourier series is a technique for representing an expression as the sum of harmonics - cosine and sinusoids. Therefore, this analysis is also known as "harmonic analysis".
Fourier series - the ideal technique before the "computer age"
>Before the creation of computer technology, the 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 complex form allows solving not only simple problems that can be directly applied to the laws of Newton's mechanics, but also fundamental equations. Most of the discoveries of Newtonian science in the nineteenth century were made possible only by Fourier's technique.
Fourier series today
>With the development of computers, Fourier transforms have risen to a qualitatively new level. This technique is firmly entrenched in almost all areas of science and technology. An example is a digital audio and video signal. Its realization became possible only thanks to the theory developed by a French mathematician at the beginning of the nineteenth century. Thus, the Fourier series in a complex form made it possible to make a breakthrough in the study of outer space. In addition, this influenced the study of the physics of semiconductor materials and plasma, microwave acoustics, oceanography, radar, and seismology.
Trigonometric Fourier series
>In mathematics, a Fourier series is a way of representing arbitrary complex functions as a sum of simpler ones. In general cases, the number of such expressions can be infinite. Moreover, the more their number is taken into account in the calculation, the more accurate the final result is. Most often, the trigonometric functions of cosine or sine are used as the simplest ones. In this case, the Fourier series are called trigonometric, and the solution of such expressions is called the expansion of the harmonic. This method plays an important role in mathematics. First of all, the trigonometric series provides a means for the image, as well as the study of functions, it is the main apparatus of the theory. In addition, it allows solving a number of problems of mathematical physics. Finally, this theory contributed to the development and brought to life a number of very important sections of mathematical science (the theory of integrals, the theory of periodic functions). In addition, it served as a starting point for the development of the following functions of a real variable, and also marked the beginning of harmonic analysis.
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. Truth.
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:
one) . 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 defined 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 out the 1 and -1 in brackets and 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)
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 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 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 No. 2. There is no escape from the spaceship - you have to decide =) An approximate design sample 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 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:
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:
Fourier series expansion of even and odd functions expansion of a function given on a segment into a series in terms of sines or cosines Fourier series for a function with an arbitrary period Complex representation of the Fourier series Fourier series in general orthogonal systems of functions Fourier series in an orthogonal system Minimal property of Fourier coefficients Bessel's inequality Equality Parseval Closed systems Completeness and closedness of systems
Fourier series expansion of even and odd functions The function f(x), defined on the segment \-1, where I > 0, is called even if the Graph of the even function is symmetrical about the y-axis. The function f(x) defined on the segment J, where I > 0, is called odd if the Graph of the odd function is symmetrical with respect to the origin. Example. a) The function is even on the segment |-jt, jt), since for all x e b) The function is odd, since the Fourier series expansion of even and odd functions is the expansion of a function given on the segment in a series of sines or cosines Fourier series for a function with an arbitrary period Complex notation of the Fourier series Fourier series in general orthogonal systems of functions Fourier series in an orthogonal system Minimum property of Fourier coefficients Bessel inequality Parseval equality Closed systems Completeness and closedness of systems c) Function f(x)=x2-x, where does not belong neither to even nor to odd functions, since Let the function f(x) satisfying the conditions of Theorem 1 be even on the segment x|. Then for all i.e. /(g) cos nx is an even function, and f(x)sinnx is an odd one. Therefore, the Fourier coefficients of an even function /(x) will be equal. Therefore, the Fourier series of an even function has the form f(x) sin nx is an even function. Therefore, we will have Thus, the Fourier series of an odd function has the form We have Applying integration by parts twice, we get that Hence, the Fourier series of this function looks like this: or, in expanded form, This equality is valid for any x €, since at the points x = ±ir the sum of the series coincides with the values of the function f(x ) = x2, since the graphs of the function f(x) = x and the sums of the resulting series are given in fig. Comment. This Fourier series allows you to find the sum of one of the convergent numerical series, namely, for x \u003d 0, we get that The function /(x) satisfies the conditions of Theorem 1, therefore it can be expanded into a Fourier series, which, due to the oddness of this function, will have the form Integrating by parts, we find the Fourier coefficients Therefore, the Fourier series of this function has the form This equality holds for all x В points x - ±tg the sum of the Fourier series does not coincide with the values of the function / (x) = x, since it is equal to Outside the segment [- *, n-] the sum of the series is a periodic continuation of the function / (x) \u003d x; its graph is shown in Fig. 6. § 6. Expansion of a function given on an interval into a series in terms of sines or cosines. Let a bounded piecewise monotonic function / be given on an interval . The values of this function on the interval 0| can be defined in various ways. For example, it is possible to define the function / on the segment mc] in such a way that /. In this case it is said that) "is extended to the segment 0] in an even way"; its Fourier series will contain only cosines. If, however, the function /(x) is defined on the segment [-x, mc] so that /(, then an odd function is obtained, and then we say that / "is extended to the segment [-*, 0] in an odd way"; in this In this case, the Fourier series will contain only sines.So, each bounded piecewise-monotone function /(x), defined on the segment , can be expanded into a Fourier series both in terms of sines and cosines.Example 1. Expand the function in a Fourier series: a) by cosines; b) along the sines. M This function, with its even and odd extensions to the segment |-x, 0) will be bounded and piecewise monotonic. a) We continue / (z) into the segment 0) a) We continue j \ x) into the segment (-m, 0 | in an even way (Fig. 7), then its Fourier series i will have the form P \u003d 1 where the Fourier coefficients are equal, respectively for Therefore, b) Let's continue /(z) in the segment [-x,0] in an odd way (Fig. 8). Then its Fourier series §7. Fourier Series for a Function with an Arbitrary Period Let the function fix) be periodic with a period of 21.1 ^ 0. To expand it into a Fourier series on the interval where I > 0, we make a change of variable by setting x = jt. Then the function F(t) = / ^tj will be a periodic function of the argument t with a period and it can be expanded on a segment in a Fourier series Returning to the variable x, i.e., setting, we obtain , remain in force also for periodic functions with an arbitrary period 21. In particular, the sufficient criterion for the expansion of a function into a Fourier series also remains valid. Example 1. Expand in a Fourier series a periodic function with a period of 21, given on the segment [-/,/] by the formula (Fig. 9). Since this function is even, its Fourier series has the form Substituting the found values of the Fourier coefficients into the Fourier series, we obtain We note one important property of periodic functions. Theorem 5. If a function has a period T and is integrable, then for any number a the equality m holds. i.e. the integral on a segment whose length is equal to the period T has the same value regardless of the position of this segment on the real axis. Indeed, We make a change of variable in the second integral, assuming This gives and therefore, Geometrically, this property means that in the case of the area shaded in Fig. 10 areas are equal to each other. In particular, for a function f(x) with a period, we obtain at the Fourier series expansion of even and odd functions the expansion of a function given on a segment into a series in terms of sines or cosines Fourier series for a function with an arbitrary period Complex representation of the Fourier series Fourier series in general orthogonal systems functions Fourier series in an orthogonal system Minimal property of Fourier coefficients Bessel inequality Parseval equality Closed systems Completeness and closedness of systems that the Fourier coefficients of a periodic function f(x) with a period of 21 can be calculated using the formulas where a is an arbitrary real number (note that the functions cos - and sin have a period of 2/). Example 3. Expand in a Fourier series a function given on an interval with a period of 2x (Fig. 11). 4 Find the Fourier coefficients of this function. Putting in the formulas we find that for Therefore, the Fourier series will look like this: At the point x = jt (discontinuity point of the first kind) we have §8. Complex notation of the Fourier series In this section, some elements of complex analysis are used (see Chapter XXX, where all the operations performed here with complex expressions are strictly justified). Let the function f(x) satisfy sufficient conditions for expansion into a Fourier series. Then on the segment x] it can be represented by a series of the form Using the Euler formulas Substituting these expressions into the series (1) instead of cos nx and sin xy we will have We introduce the following notation Then the series (2) takes the form Thus, the Fourier series (1) is presented in the complex form (3). Let us find expressions for the coefficients in terms of integrals. We have Similarly, we find Finally, the formulas for с„, с_п and с can be written as follows: . . The coefficients cn are called the complex Fourier coefficients of the function For a periodic function with a period), the complex form of the Fourier series takes the form values w if limits exist Example. Expand the period function into a complex Fourier series This function satisfies sufficient conditions for expansion into a Fourier series. Let Find the complex Fourier coefficients of this function. We have for odd for even n, or, in short. Substituting the values), we finally obtain Note that this series can also be written as follows: Fourier series in general orthogonal systems of functions 9.1. Orthogonal Systems of Functions Denote by the set of all (real) functions that are square-defined and integrable on the interval [a, 6], i.e., those for which there exists an integral. In particular, all functions f(x) that are continuous on the interval [a , 6], belong to 6], and the values of their Lebesgue integrals coincide with the values of the Riemann integrals. Definition. The system of functions, where, is called orthogonal on the interval [a, b\, if Condition (1) assumes, in particular, that none of the functions is identically equal to zero. The integral is understood in the sense of Lebesgue. and we call the quantity the norm of a function. If in an orthogonal system for any n we have, then the system of functions is called orthonormal. If the system (y>n(x)) is orthogonal, then the system Example 1. A trigonometric system is orthogonal on a segment. The system of functions is an orthonormal system of functions on, Example 2. The cosine system and the sine system is orthonormal. Let us introduce the notation that they are orthogonal on the segment (0, f|, but not orthonormal (for I ↦ 2). Since their norms are COS that the functions form an orthonormal system of functions on a segment. Let us show, for example, that the Legendre polynomials are orthogonal. Let m > n. In this case, integrating n times by parts, we find, since for the function t/m = (z2 - I)m, all derivatives up to order m - I inclusive vanish at the ends of the segment [-1,1). Definition. The system of functions (pn(x)) is called orthogonal on the interval (a, b) by overhang p(x) if: 1) there are integrals for all n = 1,2,... Here it is assumed that the weight function p(x) is defined and positive everywhere on the interval (a, b), with the possible exception of a finite number of points where p(x) can vanish. After performing differentiation in formula (3), we find. It can be shown that the Chebyshev-Hermite polynomials are orthogonal on the interval Example 4. The system of Bessel functions (jL(pix)^ is orthogonal on the interval of zeros of the Bessel function system Let an orthogonal system of functions in the interval (a, 6) and let the series (cj = const) converge on this interval to the function f(x): By virtue of the orthogonality of the system, we obtain that This operation has, generally speaking, a purely formal character. However, in some cases, for example, when the series (4) converges uniformly, all functions are continuous and the interval (a, 6) is finite, this operation is legal. But it is the formal interpretation that is important for us now. So let's say a function is given. We form the numbers c * according to the formula (5) and write The series on the right side is called the Fourier series of the function f (x) with respect to the system (^n (n)) - The numbers Cn are called the Fourier coefficients of the function f (x) in this system. The sign ~ in formula (6) only means that the numbers Cn are related to the function f(x) by formula (5) (in this case, it is not assumed that the series on the right converges at all, much less converges to the function f(x)). Therefore, the question naturally arises: what are the properties of this series? In what sense does it "represent" the function f(x)? 9.3. Average Convergence Definition. A sequence converges to an element ] on average if the norm is in space Theorem 6. If a sequence ) converges uniformly, then it also converges on average. M Let the sequence ()) converge uniformly on the segment [a, b] to the function f(x). This means that for any, for all sufficiently large n, we have Hence, from which our assertion follows. The converse is not true: the sequence () can converge on average to /(x), but not be uniformly convergent. Example. Let us consider the sequence nx It is easy to see that But this convergence is not uniform: there exists e, for example, such that no matter how large n is, on the segment Fourier series for a function with an arbitrary period Complex representation of the Fourier series Fourier series in general orthogonal systems of functions Fourier series in an orthogonal system Minimal property of Fourier coefficients Bessel inequality Parseval equality Closed systems Completeness and closedness of systems and let ) in the orthonormal system b Consider a linear combination where n ^ 1 is a fixed integer, and find the values of the constants for which the integral takes its minimum value. Let us write it in more detail Integrating term by term, due to the orthonormality of the system, we obtain The first two terms on the right side of equality (7) are independent, and the third term is nonnegative. Therefore, the integral (*) takes on a minimum value at ak = sk. The integral is called the root-mean-square approximation of the function f(x) as a linear combination of Tn(x). Thus, the root-mean-square approximation of the function /\ takes on a minimum value when. when Tn(x) is the 71st partial sum of the Fourier series of the function /(x) in the system (. Setting ak = ck, from (7) we obtain Equality (9) is called the Bessel identity. Since its left side is non-negative, then from it Bessel's inequality follows Since i is arbitrary here, Bessel's inequality can be represented in a strengthened form, i.e., for any function /, the series of squared Fourier coefficients of this function in an orthonormal system ) converges. Since the system is orthonormal on the segment [-x, r], then inequality (10) translated into the usual notation of the trigonometric Fourier series gives the relation do valid for any function f(x) with an integrable square. If f2(x) is integrable, then, by virtue of the necessary condition for the convergence of the series on the left side of inequality (11), we obtain that. Parseval's equality For some systems (^n(x)) the inequality sign in formula (10) can be replaced (for all functions f(x) 6 x) by an equals sign. The resulting equality is called the Parseval-Steklov equality (completeness condition). The Bessel identity (9) allows us to write condition (12) in an equivalent form by the space norm 6]. Definition. An orthonormal system ( is called complete in b2[ay b] if any function can be approximated with any accuracy on the average by a linear combination of the form with a sufficiently large number of terms, i.e. if for any function f(x) ∈ b2[a, b\ and for any e > 0 there is a natural number nq and numbers a\, a2y..., such that No. Theorem 7. If the system ) is complete in space by orthonormalization, the Fourier series of any function / in this system converges to f( x) on the average, i.e., according to the norm It can be shown that the trigonometric system is complete in space. This implies the assertion. Theorem 8. If a function /0 its trigonometric Fourier series converges to it on the average. 9.5. closed systems. Completeness and closedness of systems Definition. An orthonormal system of functions \, is called closed if in the space Li\a, b) there is no non-zero function orthogonal to all functions. In the space L2\a, b\ the concepts of completeness and closedness of orthonormal systems coincide. Exercises 1. Expand the function in the Fourier series in the interval (-i-, x) 2. Expand the function in the Fourier series in the interval (-r, r) 3. Expand the function in the Fourier series in the interval (-r, r) 4. Expand in a Fourier series in the interval (-jt, r) function 5. Expand in a Fourier series in the interval (-r, r) the function f(x) = x + x. 6. Expand in a Fourier series in the interval (-jt, r) the function n 7. Expand in a Fourier series in the interval (-r, x) the function / (x) \u003d sin2 x. 8. Expand in a Fourier series in the interval (-m, jt) the function f(x) = y 9. Expand in a Fourier series in the interval (-mm, -k) the function f(x) = | sinx|. 10. Expand in a Fourier series in the interval (-x-, r) the function f(x) = g. 11. Expand in a Fourier series in the interval (-r, r) the function f (x) \u003d sin §. 12. Expand in a Fourier series the function f (x) = n -2x, given in the interval (0, x), continuing it in the interval (-x, 0): a) in an even way; b) in an odd way. 13. Expand in a Fourier series in terms of sines the function / (x) \u003d x2, given in the interval (0, x). 14. Expand in a Fourier series the function / (x) \u003d 3-x, given in the interval (-2,2). 15. Expand in a Fourier series the function f (x) \u003d |x |, given in the interval (-1,1). 16. Expand in a Fourier series in terms of sines the function f (x) \u003d 2x, specified in the interval (0,1).
Fourier series- a way of representing a complex function as a sum of simpler, well-known ones.
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 YY Y on the coordinate plane. In the same way that we can describe the coordinates of a point with respect to the axes, we can describe any function with respect to sines and cosines. Trigonometric functions are well understood and easy to apply in mathematics.
You can represent sines and cosines in the form of such waves:
Blue are cosines, red are sines. These waves are also called harmonics. Cosines are even, sines are odd. The term harmonica comes from antiquity and is associated with observations about the relationship of pitches in music.
What is a Fourier Series
>Such a series, where the sine and cosine functions are used as the simplest, is called trigonometric. It is named after its inventor Jean Baptiste Joseph Fourier, at the end of the 18th – beginning of the 19th century. who proved that any function can be represented as a combination of such harmonics. And the more you take, the more accurate this representation will be. For example, the picture below: you can see that with a large number of harmonics, i.e., members of the Fourier series, the red graph gets closer to the blue one - the original function.
Practical application in the modern world
>Are these rows really needed now? Where can they be applied in practice and does anyone other than theoretical mathematicians use them? It turns out that Fourier is famous all over the world because the practical use of his series is literally incalculable. It is convenient to use them where there are any vibrations or waves: acoustics, astronomy, radio engineering, etc. The simplest example of its use is the mechanism of the camera or video camera. In short, these devices record not just pictures, but the coefficients of Fourier series. And it works everywhere - when viewing pictures on the Internet, a movie or listening to music. It is thanks to Fourier series that you can now read this article from your mobile phone. Without the Fourier transform, we would not have enough bandwidth of Internet connections to simply watch a YouTube video, even in standard quality.
In this diagram, the two-dimensional Fourier transform, which is used to decompose the image into harmonics, i.e., basic components. In this diagram, the value -1 is encoded in black, 1 in white. To the right and down the graph, the frequency increases.
Fourier expansion
>Probably, you are already tired of reading, so let's move on to the formulas.
For such a mathematical technique as the expansion of functions in a Fourier series, one will have to take integrals. Lots of integrals. In general, the Fourier series is written as an infinite sum:
F (x) = A + ∑ n = 1 ∞ (a n cos (n x) + b n sin (n x)) f(x) = A + \displaystyle\sum_(n=1)^(\infty)(a_n \cos(nx)+b_n \sin(nx))f(x) =A+n=1∑ ∞ (a n cos (n x ) +b n sin (n x ) )
where
A = 1 2 π ∫ − π π f (x) d x A = \frac(1)(2\pi)\displaystyle\int\limits_(-\pi)^(\pi) f(x)dxA=2 pi1 − π ∫ π f(x)dx
a n = 1 π ∫ − π π f (x) cos (n x) d x a_n = \frac(1)(\pi)\displaystyle\int\limits_(-\pi)^(\pi) f(x)\ cos(nx)dxa n = π 1 − π ∫ π f(x)cos(nx)dx
b n = 1 π ∫ − π π f (x) sin (n x) d x b_n = \frac(1)(\pi)\displaystyle\int\limits_(-\pi)^(\pi) f(x)\ sin(nx)dxb n = π 1 − π ∫ π f(x)sin(nx)dx
If we can somehow count an infinite number of a n a_n a n and b n b_n b n (they are called the coefficients of the Fourier expansion, A A A is just a constant of this expansion), then the resulting series will 100% coincide with the original function f(x) f(x) f(x) on the segment from − π -\pi − π before π\pi π . Such a segment is due to the integration properties of sine and cosine. The more n n n, for which we calculate the coefficients of the expansion of the function into a series, the more accurate this expansion will be.
ExampleLet's take a simple function y=5x y=5x y=5 x
A = 1 2 π ∫ − π π f (x) d x = 1 2 π ∫ − π π 5 x d x = 0 A = \frac(1)(2\pi)\displaystyle\int\limits_(-\pi)^ (\pi) f(x)dx = \frac(1)(2\pi)\displaystyle\int\limits_(-\pi)^(\pi) 5xdx = 0A=2 pi1
−
π
∫
π
f (x) d x =2 pi1
−
π
∫
π
5xdx=0
a 1 = 1 π ∫ − π π f (x) cos (x) d x = 1 π ∫ − π π 5 x cos (x) d x = 0 a_1 = \frac(1)(\pi)\displaystyle\ int\limits_(-\pi)^(\pi) f(x)\cos(x)dx = \frac(1)(\pi)\displaystyle\int\limits_(-\pi)^(\pi) 5x \cos(x)dx = 0a 1
=
π
1
−
π
∫
π
f (x ) cos (x ) d x =π
1
−
π
∫
π
5xcos(x)dx=0
b 1 = 1 π ∫ − π π f (x) sin (x) d x = 1 π ∫ − π π 5 x sin (x) d x = 10 b_1 = \frac(1)(\pi)\displaystyle\ int\limits_(-\pi)^(\pi) f(x)\sin(x)dx = \frac(1)(\pi)\displaystyle\int\limits_(-\pi)^(\pi) 5x \sin(x)dx = 10b 1
=
π
1
−
π
∫
π
f (x) sin (x) d x =π
1
−
π
∫
π
5xsin(x)dx=1
0
a 2 = 1 π ∫ − π π f (x) cos (2 x) d x = 1 π ∫ − π π 5 x cos (2 x) d x = 0 a_2 = \frac(1)(\pi)\ displaystyle\int\limits_(-\pi)^(\pi) f(x)\cos(2x)dx = \frac(1)(\pi)\displaystyle\int\limits_(-\pi)^(\pi ) 5x\cos(2x)dx = 0a 2
=
π
1
−
π
∫
π
f (x ) cos (2 x ) d x =π
1
−
π
∫
π
5 x cos (2 x ) d x =0
b 2 = 1 π ∫ − π π f (x) sin (2 x) d x = 1 π ∫ − π π 5 x sin (2 x) d x = − 5 b_2 = \frac(1)(\pi) \displaystyle\int\limits_(-\pi)^(\pi) f(x)\sin(2x)dx = \frac(1)(\pi)\displaystyle\int\limits_(-\pi)^(\ pi) 5x\sin(2x)dx = -5b 2
=
π
1
−
π
∫
π
f(x)
sin(2
x)
dx=
π
1
−
π
∫
π
5
xsin(2
x)
dx=
−
5
And so on. In the case of such a function, we can immediately say that all a n = 0 a_n=0
5 x ≈ 10 ⋅ sin (x) − 5 ⋅ sin (2 ⋅ x) + 10 3 ⋅ sin (3 ⋅ x) − 5 2 ⋅ sin (4 ⋅ 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 \ cdotx)
The graph of the resulting function will look like this:
The resulting Fourier expansion approaches our original function. If we take a larger number of terms in the series, for example, 15, we will already see the following:
The more expansion terms in a series, the higher the accuracy.
If we change the scale of the graph a little, we can notice another feature of the transformation: the Fourier series is a periodic function with a period 2 π 2\pi
Thus, it is possible to represent any function that is continuous on the interval [ − π ; pi ] [-\pi;\pi]
