Fourier series. For each day Expand the function in a Fourier series

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

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

Appointment. The online calculator is designed to expand the function f(x) in a Fourier series.

For modulo functions (e.g. |x|), use cosine expansion.

Function entry rules:

For modulo functions, use the cosine expansion. For example, for |x| it is necessary to introduce a function without a module, i.e. x .

Fourier series piecewise-continuous, piecewise-monotone and bounded on the interval (- l;l) of the function converges on the entire real axis.

The sum of the Fourier series S(x) :

• is a periodic function with period 2 l. A function u(x) is called periodic with period T (or T-periodic) if for all x of the domain R, u(x+T)=u(x).
• on the interval (- l;l) coincides with the function f(x), except for break points
• at discontinuity points (of the first kind, since the function is limited) of the function f(x) and takes average values ​​at the ends of the interval:

.
They say that the function expands into a Fourier series on the interval (- l;l): .

If a f(x) is an even function, then only even functions participate in its expansion, that is, b n=0.
If a f(x) is an odd function, then only odd functions participate in its expansion, that is, a n=0

Near Fourier functions f(x) on the interval (0; l) by cosines of multiple arcs the row is called:
, where
.
Near Fourier functions f(x) on the interval (0; l) by sines of multiple arcs the row is called:
, where .
The sum of the Fourier series over the cosines of multiple arcs is an even periodic function with period 2 l, coinciding with f(x) on the interval (0; l) at points of continuity.
The sum of the Fourier series over the sines of multiple arcs is an odd periodic function with a period of 2 l, coinciding with f(x) on the interval (0; l) at points of continuity.
The Fourier series for a given function on a given interval has the property of uniqueness, that is, if the expansion is obtained in any other way than using formulas, for example, by selecting coefficients, then these coefficients coincide with those calculated by the formulas.

Example #1. Expand the function f(x)=1:
a) in a complete Fourier series on the interval(-π ;π);
b) in a series along the sines of multiple arcs on the interval(0;π); plot the resulting Fourier series
Solution:
a) The expansion in the Fourier series on the interval (-π; π) has the form:
,
and all the coefficients b n=0, because this function is even; thus,

Obviously, the equality will be satisfied if we take
a 0 =2, a 1 =a 2 =a 3 =…=0
By virtue of the uniqueness property, these are the desired coefficients. Thus, the required expansion is: or just 1=1.
In this case, when the series coincides identically with its function, the graph of the Fourier series coincides with the graph of the function on the entire real line.
b) The expansion on the interval (0;π) in terms of the sines of multiple arcs has the form:
It is obviously impossible to choose the coefficients so that the equality holds identically. Let's use the formula to calculate the coefficients:


Thus, for even n (n=2k) we have b n=0, for odd ( n=2k-1) -
Finally, .
Let's plot the resulting Fourier series using its properties (see above).
First of all, we build a graph of this function on a given interval. Further, taking advantage of the oddness of the sum of the series, we continue the graph symmetrically to the origin:

We continue in a periodic way on the entire number axis:


And finally, at the break points, we fill in the average (between the right and left limits) values:

Example #2. Expand function on the interval (0;6) along the sines of multiple arcs.
Solution: The desired expansion has the form:

Since both the left and right parts of the equality contain only sin functions of different arguments, you should check whether the arguments of the sines in the left and right parts of the equality coincide for any values ​​of n (natural!)
or , whence n =18. This means that such a term is contained in the right side and the coefficient for it must coincide with the coefficient on the left side: b 18 =1;
or , whence n =4. Means, b 4 =-5.
Thus, using the selection of coefficients, it was possible to obtain the desired expansion.

Function defined for all values x called periodical, if there is such a number T (T≠ 0), that for any value x equality f(x + T) = f(x). Number T in this case is the period of the function.

Properties of periodic functions:

1) Sum, difference, product and quotient of periodic period functions T is a periodic function of the period T.

2) If the function f(x) has a period T, then the function f(ax) has a period

Indeed, for any argument X:

(multiplying the argument by a number means squeezing or stretching the graph of this function along the axis OH)

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

3) If f(x) period periodic function T, then any two integrals of this function are equal, taken over the interval of length T(it is assumed that these integrals exist).

Fourier series for a function with period T= .

A trigonometric series is a series of the form:

or, in short,

Where , , , , , … , , , … are real numbers, called coefficients of the series.

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

period, and the period () is , and hence ). Each term (), with n= 1,2,3… is an analytical expression of a simple harmonic oscillation , where A- amplitude,

initial phase. Given the above, we get: if the trigonometric series converges on a segment of the length of the period, then it converges on the entire numerical axis and its sum is a periodic function of the period.

Let the trigonometric series converge uniformly on a segment (and therefore on any segment) and its sum is equal to . To determine the coefficients of this series, we use the following equalities:

We also use the following properties.

1) As is known, the sum of a series composed of continuous functions uniformly convergent on a certain segment is itself a continuous function on this segment. Taking this into account, we obtain that the sum of a trigonometric series uniformly converging on a segment is a continuous function on the entire real axis.

2) The uniform convergence of the series on a segment will not be violated if all terms of the series are multiplied by a function that is continuous on this segment.

In particular, uniform convergence on a segment of a given trigonometric series will not be violated if all members of the series are multiplied by or by .

By condition

As a result of term-by-term integration of the uniformly convergent series (4.2) and taking into account the above equalities (4.1) (orthogonality of trigonometric functions), we obtain:

Therefore, the coefficient

Multiplying equality (4.2) by , integrating this equality within the range from to and, taking into account the above expressions (4.1), we obtain:

Therefore, the coefficient

Similarly, multiplying equality (4.2) by and integrating it within the limits from to , taking equalities (4.1) into account, we have:

Therefore, the coefficient

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

Sufficient criteria for the expansion of a function into a Fourier series. Recall that the point x o function break f(x) is called a discontinuity point of the first kind if there are finite limits on the right and left of the function f(x) in the vicinity of the point.

Limit on the right

Left limit.

Theorem (Dirichlet). If the function f(x) has a period and is continuous on the segment or has a finite number of discontinuity points of the first kind and, in addition, the segment can be divided into a finite number of segments so that inside each of them f(x) is monotonic, then the Fourier series for the function f(x) converges for all values x. Moreover, at the points of continuity of the function f(x) its sum is f(x), and at the discontinuity points of the function f(x) its sum is , i.e. the arithmetic mean of the limit values ​​on the left and right. In addition, the Fourier series for the function f(x) converges uniformly on any segment that, together with its ends, belongs to the interval of continuity of the function f(x).

Example: expand the function in a Fourier series

Satisfying the condition.

Solution. Function f(x) satisfies the Fourier expansion conditions, so we can write:

In accordance with formulas (4.3), one can obtain the following values ​​of the coefficients of the Fourier series:

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

And therefore

Fourier series for even and odd functions with period T = .

We use the following property of the integral over a symmetric with respect to x=0 span:

If a f(x)- odd function,

if f(x) is an even function.

Note that the product of two even or two odd functions is an even function, and the product of an even function and an odd function is an odd function. Let now f(x)- even periodic function with period , which satisfies the conditions of expansion into a Fourier series. Then, using the above property of integrals, we get:

Thus, the Fourier series for an even function contains only even functions - cosines and is written as follows:

and the coefficients bn = 0.

Arguing similarly, we get that if f(x) - an odd periodic function that satisfies the conditions of expansion into a Fourier series, then, therefore, the Fourier series for an odd function contains only odd functions - sines and is written as follows:

wherein an=0 at n=0, 1,…

Example: expand in a Fourier series a periodic function

Since the given odd function f(x) satisfies the Fourier expansion conditions, then

or, which is the same,

And the Fourier series for this function f(x) can be written like this:

Fourier series for functions of any period T=2 l.

Let f(x)- periodic function of any period T=2l(l- half-period), piecewise-smooth or piecewise-monotone on the interval [ -l,l]. Assuming x=at, get the function f(at) argument t, whose period is . Let's pick a so that the period of the function f(at) was equal to , i.e. T = 2l

Solution. Function f(x)- odd, satisfying the conditions of expansion into a Fourier series, therefore, based on formulas (4.12) and (4.13), we have:

(when calculating the integral, the formula "integration by parts" was used).

follows:

1) draw a graph f(x) on an interval of at least two periods long, to show that the given function is periodic;

2) draw a graph S(x) similarly, so that it can be seen at what points f(x)¹S(x);

3) calculate the Fourier coefficients and write down the Fourier series.

Tasks

№1. Expand in a Fourier Series

Solution. notice, that f(x) given on the length interval T=4. Because f(x) is assumed to be periodic, then it is this number that is its period, then - l = 2.

1) Graph f(x):

2) Graph S(x):

The arrows at the ends of the lines show that the function does not take at the ends of the interval the value determined from the expression given on the interval. When comparing graphs f(x) and S(x) it is clearly seen that at the discontinuity points f(x)¹S(x).

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

Decomposition f(x) in a Fourier series has the form:

Remarks . 1) When integrating on [-1;3] this section has been divided into and , because on these segments f(x) set to different values.

2) When calculating the coefficients, integrals were used: and , where a = const.

№2 . Expand in a Fourier Series

Solution. Here T=2, l = 1.

The Fourier series has the form: , where ; ; , because l = 1.

1) Graph f(x):

2) Graph S(x):

№3. Expand in a Fourier series in terms of sines

Solution. Note that only odd functions are expanded in the Fourier series in terms of sines. Because f(x) defined only for x > 0, xн(0;2)И(2;3), then this means that on the symmetric interval (-3;-2)È(-2;0) f(x) must be continued in such a way that the equality f(-x) = -f(x). Therefore, the length of the interval on which f(x) given as an odd function, equals 6. Hence T = 6, l = 3. Fourier series for f(x) has the form: , where , n = 1, 2, 3, (according to formulas (5")).

1) Graph f(x).

To draw a graph f(x) as an odd function, we first draw a graph on (0;2)È(2;3), and then take advantage of the fact that the graph of an odd function is symmetrical with respect to the origin. From these considerations, we get the graph f(x) on the (-3;-2)È(-2;0). Then we continue f(x) T=6.

2) Graph S(x).

Schedule S(x) different from chart f(x) at the break points of the function f(x). For example, in t. x = 2f(x) not defined, but S(x) has at x=2 a value equal to half the sum of the one-sided limits of the function f(x), exactly: , where , .

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

№4 . Expand in a Fourier series in cosines.

Solution. Note that only even functions can be expanded in the Fourier series in cosines. Because f(x) set only for x>0, xн(0;2)И(2;3], then this means that on the symmetric interval [-3;-2)È(-2;0) f(x) we need to continue in such a way that the equality holds: f(-x) = f(x). Therefore, the length of the interval on which f(x) given as an even function is equal to 6, then T = 6, l = 3. The Fourier series in this case has the form:

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

1) Graph f(x).

To draw a graph f(x) as an even function, we first draw a graph f(x) on the (0;2)È(2;3], and then take advantage of the fact that the graph of an even function is symmetrical about the y-axis. From these considerations, we get the graph f(x) on the [-3;-2)È(-2;0). Then we continue f(x) on the entire number line as a periodic function with period T=6.

Here is the chart f(x) drawn on two full periods of the function.

2) Graph S(x).

Schedule S(x) different from chart f(x) at the break points of the function f(x). For example, in t. x = 0 f(x) not defined, but S(x) has the meaning: , so the graph S(x) is not interrupted in x=0, in contrast to the graph f(x).

Decomposition f(x) in a Fourier series in cosines has the form: .

№5. Expand in a Fourier Series f(x) = |x|, xн(-2;2)..

Solution. By condition, f(x) is an even function on (-2;2) ; those. its Fourier series contains only cosines, while T = 4, l = 2, ,

where ; ; n = 1, 2,

1) Graph f(x):

2) Graph S(x):

3) , because |x| = x for x > 0.; .

Then the decomposition f(x) in a Fourier series has the form: . Note that when integrating expressions or , the integration-by-parts formula is used: , where u=x; dv = cos(ax)dx or dv = sin(ax)dx.

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

Solution. a) Graph of a function with 2? - periodic continuation has the form

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

Let us calculate the Fourier coefficients. Since the function is even, then bn = 0 (n = 0, 1, 2,…) and (n = 0, 1, 2,…).

To calculate this integral, the formula for integration by parts in a definite integral is used. We get

The Fourier series of this function has the form . By virtue of the Dirichlet test, this series represents the function x2 in the interval (-?,?).

b) The interval (0, 2?) is not symmetrical with respect to the origin, and its length is 2 l= 2?. We calculate the Fourier coefficients using the formulas:

Therefore, the Fourier series has the form . By virtue of the Dirichlet theorem, the series converges to a generating function at the points x?(0,2?), and at the points 0 and 2? to value. The series sum graph looks like

c) The function expanded in a series in terms of sines must be odd. Therefore, we extend the given function x2 in (-π,π) in an odd way, i.e. consider the function. For this function f(x) we have an = 0 (n = 0, 1, 2,…) and

The desired expansion has the form .

The series sum graph looks like

Note that at the points x = (-π, π) the Fourier series converges to zero.

№7 Expand in a Fourier series a function given graphically:

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

This function satisfies the conditions of the Dirichlet test, so it expands into a Fourier series. Let us calculate the Fourier coefficients ( l = 1):

; (n = 1, 2,…);

The Fourier series for the function f(x) has the form

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

№8. Expand the function into a trigonometric Fourier series on a segment and indicate the function to which the resulting series converges.

Solution. Draw a graph of a function, continuing it periodically with a period or on the entire axis. The continued function has a period.

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

The function is piecewise monotonic on the segment : it increases on and on . At points, the function has discontinuities of the first kind.

Find out if a function is even or odd: The function is neither even nor odd.

a) if the function is set to

b) if the function is set to

Compose the Fourier series of the function: .

Specify the function to which this series will converge, using pointwise convergence criteria: According to the Dirichlet criterion, the Fourier series of the function converges to the sum:

№9. Expand the function into a Fourier series in terms of sines on and use this expansion to find the sum of the number series.

Solution. Continue the function in an even (odd) way on (- p,0) or (- l,0), and then periodically with period 2 p or 2 l continue the function to the whole axis.

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

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

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

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

Calculate Fourier coefficients:

Its Fourier coefficients are calculated by the formulas:

Compose the Fourier series of the function. .

Specify the function to which this series will converge, using pointwise convergence criteria.

According to the Dirichlet test, the Fourier series of the function converges to the sum:

Therefore, when

Substituting the values, indicate the sum of the given number series.

Assuming in the resulting decomposition , we find ,

whence, since , .

№10. Write Parseval's equality for the function , and, based on this equality, find the sum of the number series .

Solution. Determine whether the given function is a square integrable function on .

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

Calculate the Fourier coefficients using the formulas:

Since it is an odd function, its Fourier coefficients are calculated by the formulas:

Calculate integral .

Write the Parseval formula:

Thus, the Parseval formula has the form

Having performed, if necessary, arithmetic operations on the right and left sides, get the sum of the given numerical series.

Dividing both parts of the resulting equality by 144, we find: .

№11. Find the Fourier Integral of a Function

and build its graph.

Solution. Plot the function.

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

The function is absolutely integrable in the interval, continuous for and , and has a discontinuity of the first kind at a point. Further, for and the function has a finite derivative, and at zero there are finite right and left derivatives. Find out if the function is even or odd. The function is neither even nor odd. ; .

So , or ,

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1 MINISTRY OF EDUCATION AND SCIENCE OF THE RUSSIAN FEDERATION NOVOSIBIRSK STATE UNIVERSITY FACULTY OF PHYSICS R. K. Belkheeva FOURIER SERIES IN EXAMPLES AND TASKS Tutorial Novosibirsk 211

2 UDC BBK V161 B44 B44 Belkheeva R. K. Fourier series in examples and problems: Textbook / Novosib. state un-t. Novosibirsk, s. ISBN The tutorial provides basic information about Fourier series, provides examples for each topic studied. An example of applying the Fourier method to solving the problem of transverse vibrations of a string is analyzed in detail. Illustrative material is given. There are tasks for independent solution. It is intended for students and teachers of the Faculty of Physics of Novosibirsk State University. Published according to the decision of the Methodological Commission of the Faculty of Physics of NSU. Reviewer Dr. phys.-math. Sciences. V. A. Aleksandrov ISBN c Novosibirsk State University, 211 c Belkheeva R. K., 211

3 1. Fourier series expansion of a 2π-periodic function Definition. The Fourier series of the function f(x) is the functional series a 2 + (a n cosnx + b n sin nx), (1) where the coefficients a n, b n are calculated by the formulas: a n = 1 π b n = 1 π f(x) cosnxdx, n = , 1,..., (2) f(x) sin nxdx, n = 1, 2,.... (3) Formulas (2) (3) are called the Euler Fourier formulas. The fact that the function f(x) corresponds to the Fourier series (1) is written as a formula f(x) a 2 + (a n cosnx + b n sin nx) (4) and they say that the right side of formula (4) is a formal series Fourier functions f(x). In other words, formula (4) means only that the coefficients a n, b n are found by formulas (2), (3). 3

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

5 Fig. 1. Graph of the function f(x) nx + π n n 2 = 2 π (1) n 1 n 2 = b n = 1 π π = 2 π f(x) cosnxdx = cos nx cos n 2 = 4 πn2, for odd n, for even n, f(x ) sin nxdx = because the function f(x) is even. We write the formal Fourier series for the function f(x): f(x) π 2 4 π k= 5 cos (2k + 1)x (2k + 1) 2.

6 Find out whether the function f(x) is piecewise smooth. Since it is continuous, we calculate only the limits (6) at the end points of the interval x = ±π and at the break point x = : and f(π h) f(π) π h π lim = lim h + h h + h = 1, f(+ h) f(+) + h () lim = lim h + h h + h f(+ h) f(+) + h lim = lim = 1, h + h h + h = 1, f(h) f () h () lim = lim = 1. h + h h + h The limits exist and are finite, hence the function is piecewise smooth. By the pointwise convergence theorem, its Fourier series converges to the number f(x) at each point, i.e., f(x) = π 2 4 π k= cos (2k + 1) + x (2k + 1) 2 = = π 2 4 (cosx + 19 π cos 3x) cos 5x (7) Figures 2 and 3 show the nature of the approximation of the partial sums of the Fourier series S n (x), where S n (x) = a n 2 + (a k coskx + b k sin kx), k=1, to the function f(x) in the interval [, π] . 6

7 Fig. Fig. 2. Graph of the function f(x) with superimposed graphs of partial sums S (x) = a 2 and S 1(x) = a 2 + a 1 cos x 3. Graph of the function f (x) with a partial sum graph superimposed on it S 99 (x) \u003d a 2 + a 1 cos x + + a 99 cos 99x 7

8 Substituting in (7) x = we get: = π 2 4 π k= 1 (2k + 1) 2, from where we find the sum of the number series: = π2 8. Knowing the sum of this series, it is easy to find the following sum We have: S = ( ) S = ()= π S, hence S = π2 6, that is, 1 n = π The sum of this famous series was first found by Leonhard Euler. It is often found in mathematical analysis and its applications. EXAMPLE 2. Draw a graph, find the Fourier series of the function given by the formula f(x) = x for x< π, предполагая, что она имеет период 2π, и вычислим суммы числовых (1) n) рядов + n= ((2n + 1,) (k k + 1) Решение. График функции f(x) приведен на рис. 4. 8

9 Fig. 4. Graph of the function f(x) The function f(x) is continuously differentiable on the interval (, π). At the points x = ±π, it has finite limits (5): f() =, f(π) = π. In addition, there are finite limits (6): f(+ h) f(+) lim = 1 and h + h f(π h) f(π +) lim = 1. h + h Hence, f(x) is piecewise smooth function. Since the function f(x) is odd, then a n =. The coefficients b n are found by integration by parts: b n = 1 π f(x) sin πnxdx= 1 [ x cosnx π πn + 1 n = 1 πn [(1)n π + (1) n π] = 2(1)n+ one. n Let us compose the formal Fourier series of the function 2(1) n+1 f(x) sin nx. n 9 cosnxdx ] =

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

11 Fig. Fig. 6. Graph of the function f(x) with the graph of the partial sum S 2 (x) superimposed on it. 7. Graph of the function f(x) with the graph of the partial sum S 3 (x) 11 superimposed on it

12 Fig. Fig. 8. Graph of the function f(x) with the graph of the partial sum S 99 (x) superimposed on it. We use the obtained Fourier series to find the sums of two numerical series. We put in (8) x = π/2. Then 2 () +... = π 2, or = n= (1) n 2n + 1 = π 4. We easily found the sum of the well-known Leibniz series. Putting x = π/3 in (8), we find () +... = π 2 3, or (1+ 1) () (k) 3π +...= 3k

13 EXAMPLE 3. Draw a graph, find the Fourier series of the function f(x) = sin x, assuming that it has a period of 2π, and 1 calculate the sum of the number series 4n 2 1. Solution. The graph of the function f(x) is shown in fig. 9. Obviously, f(x) = sin x is a continuous even function with period π. But 2π is also the period of the function f(x). Rice. 9. Graph of the function f(x) Let's calculate the Fourier coefficients. All b n = because the function is even. Using trigonometric formulas, we calculate a n for n 1: a n = 1 π = 1 π sin x cosnxdx = 2 π sin x cosnxdx = (sin(1 + n)x sin(1 n)x) dx = = 1 () π cos( 1 + n)x cos(1 n)x + = 2 () 1 + (1) n = π 1 + n 1 n π 1 n 2 ( 4 1 if n = 2k, = π n 2 1 if n = 2k

14 This calculation does not allow us to find the coefficient a 1 because at n = 1 the denominator goes to zero. Therefore, we calculate the coefficient a 1 directly: a 1 = 1 π sin x cosxdx =. Since f(x) is continuously differentiable on (,) and (, π) and at the points kπ, (k is an integer), there are finite limits (5) and (6), the Fourier series of the function converges to it at every point: = 2 π 4 π sinx = 2 π 4 π cos 2nx 4n 2 1 = (1 1 cos 2x cos 4x + 1) cos 6x 1. Graph of the function f(x) with the graph of the partial sum S(x) superimposed on it 14

15 Fig. Fig. 11. Graph of the function f(x) with the graph of the partial sum S 1 (x) superimposed on it. Fig. 12. Graph of the function f(x) with the graph of the partial sum S 2 (x) superimposed on it. 13. Graph of the function f(x) with the graph of the partial sum S 99 (x) 15 superimposed on it

16 1 Calculate the sum of the number series. To do this, we put 4n 2 1 in (9) x =. Then cosnx = 1 for all n = 1, 2,... and Therefore, 2 π 4 π 1 4n 2 1 =. 1 4n 2 1 = = 1 2. EXAMPLE 4. Let us prove that if a piecewise smooth continuous function f(x) satisfies the condition f(x π) = f(x) for all x (i.e., is π-periodic) , then a 2n 1 = b 2n 1 = for all n 1, and vice versa, if a 2n 1 = b 2n 1 = for all n 1, then f(x) is π-periodic. Solution. Let the function f(x) be π-periodic. Let us calculate its Fourier coefficients a 2n 1 and b 2n 1: = 1 π (a 2n 1 = 1 π f(x) cos(2n 1)xdx + f(x) cos(2n 1)xdx =) f(x) cos (2n 1)xdx. In the first integral we make the change of variable x = t π : f(x) cos(2n 1)xdx = f(t π) cos(2n 1)(t + π) dt. 16

17 Using the fact that cos(2n 1)(t + π) = cos(2n 1)t and f(t π) = f(t), we get: a 2n 1 = 1 π (f(x) cos(2n 1)x dx+) f(x) cos(2n 1)x dx =. It is proved similarly that b 2n 1 =. Conversely, let a 2n 1 = b 2n 1 =. Since the function f(x) is continuous, then, by the theorem on the representability of a function at a point by its Fourier series, we have Then f(x π) = f(x) = (a 2n cos 2nx + b 2n sin 2nx). (a2n cos 2n(x π) + b 2n sin 2n(x π)) = (a2n cos 2nx + b 2n sin 2nx) = f(x), which means that f(x) is a π-periodic function. EXAMPLE 5. Let us prove that if a piecewise smooth function f(x) satisfies the condition f(x) = f(x) for all x, then a = and a 2n = b 2n = for all n 1, and vice versa, if a = a 2n = b 2n =, then f(x π) = f(x) for all x. Solution. Let the function f(x) satisfy the condition f(x π) = f(x). Let us calculate its Fourier coefficients: 17

18 = 1 π (a n = 1 π f(x) cos nxdx + f(x) cosnxdx =) f(x) cosnxdx. In the first integral we make the change of variable x = t π. Then f(x) cosnxdx = f(t π) cosn(t π) dt. Using the fact that cos n(t π) = (1) n cosnt and f(t π) = f(t), we obtain: a n = 1 π ((1) n) f(t) cosnt dt = if n even, = 2 π f(t) cos nt dt, if n is odd. π It is proved similarly that b 2n =. Conversely, let a = a 2n = b 2n =, for all n 1. Since the function f(x) is continuous, then, by the theorem on the representability of a function at a point, its Fourier series satisfies the equality f(x) = (a 2n 1 cos ( 2n 1)x + b 2n 1 sin (2n 1)x). eighteen

19 Then = f(x π) = = = f(x). EXAMPLE 6. Let us study how to extend the function f(x) integrable on the interval [, π/2] to the interval [, π], so that its Fourier series has the form: a 2n 1 cos(2n 1)x. (1) Solution. Let the graph of the function have the form shown in Fig. 14. Since in series (1) a = a 2n = b 2n = for all n, it follows from Example 5 that the function f(x) must satisfy the equality f(x π) = f(x) for all x. This observation gives a way to extend the function f(x) to the interval [, /2] : f(x) = f(x+π), fig. 15. From the fact that series (1) contains only cosines, we conclude that the continued function f (x) must be even (i.e., its graph must be symmetrical about the Oy axis), Fig.

20 Fig. 14. Graph of the function f(x) 15. Graph of the continuation of the function f(x) on the interval [, /2] 2

21 So, the desired function has the form shown in fig. 16. Fig. 16. Graph of the continuation of the function f(x) on the interval [, π] Summing up, we conclude that the function should be continued as follows: f(x) = f(x), f(π x) = f(x), that is interval [π/2, π], the graph of the function f(x) is centrally symmetric about the point (π/2,), and on the interval [, π], its graph is symmetric about the Oy axis. 21

22 GENERALIZATION OF EXAMPLES 3 6 Let l >. Consider two conditions: a) f(l x) = f(x); b) f(l + x) = f(x), x [, l/2]. From a geometric point of view, condition (a) means that the graph of the function f(x) is symmetric about the vertical line x = l/2, and condition (b) that the graph f(x) is centrally symmetric about the point (l/2;) on the axis abscissa. Then the following statements are true: 1) if the function f(x) is even and condition (a) is satisfied, then b 1 = b 2 = b 3 =... =, a 1 = a 3 = a 5 =... = ; 2) if the function f(x) is even and condition (b) is satisfied, then b 1 = b 2 = b 3 =... =, a = a 2 = a 4 =... = ; 3) if the function f(x) is odd and condition (a) is satisfied, then a = a 1 = a 2 =... =, b 2 = b 4 = b 6 =... = ; 4) if the function f(x) is odd and condition (b) is satisfied, then a = a 1 = a 2 =... =, b 1 = b 3 = b 5 =... =. PROBLEMS In problems 1 7 draw graphs and find the Fourier series for the functions, (assuming they have a period of 2π: if< x <, 1. f(x) = 1, если < x < π. 1, если < x < /2, 2. f(x) =, если /2 < x < π/2, 1, если π/2 < x < π. 3. f(x) = x 2 (< x < π). 4. f(x) = x 3 (< x < π). { π/2 + x, если < x <, 5. f(x) = π/2 x, если < x < π. 22

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

24 2. Expansion of a function given in the interval [, π] only in terms of sines or only in terms of cosines Let a function f be given in the interval [, π]. Wanting to expand it in this interval into a Fourier series, we first extend f into the interval [, π] in an arbitrary way, and then we use the Euler Fourier formulas. The arbitrariness in the continuation of a function leads to the fact that for the same function f: [, π] R we can obtain different Fourier series. But it is possible to use this arbitrariness in such a way as to obtain an expansion only in sines or only in cosines: in the first case, it suffices to continue f in an odd way, and in the second, in an even way. Solution algorithm 1. Continue the function in an odd (even) way on (,), and then periodically with a period of 2π continue the function to the entire axis. 2. Calculate the Fourier coefficients. 3. Compose the Fourier series of the function f(x). 4. Check the conditions for the convergence of the series. 5. Specify the function to which this series will converge. EXAMPLE 7. Expand the function f(x) = cosx,< x < π, в ряд Фурье только по синусам. Решение. Продолжим функцию нечетным образом на (,) (т. е. так, чтобы равенство f(x) = f(x) выполнялось для всех x (, π)), а затем периодически с периодом 2π на всю ось. Получим функцию f (x), график которой приведен на рис

25 Fig. 17. Graph of the continued function Obviously, the function f (x) is piecewise smooth. Let's calculate the Fourier coefficients: a n = for all n because the function f (x) is odd. If n 1, then b n = 2 π f(x) sin πnxdx = 2 π cosx sin nxdx = = 2 π dx = = 2 π cos (n + 1) x cos (n 1) x + = π n + 1 n 1 = 1 (1) n (1)n 1 1 = π n + 1 n 1 = 1 if n = 2 k + 1, (1)n+1 (n 1) + (n + 1) = π ( n + 1)(n 1) 2 2n if n = 2k. π n 2 1 For n = 1 in the previous calculations, the denominator vanishes, so the coefficient b 1 can be calculated directly.

26 Essentially: b 1 = 2 π cosx sin xdx =. Compose the Fourier series of the function f (x) : f (x) 8 π k=1 k 4k 2 1 sin 2kx. Since the function f (x) is piecewise smooth, then, by the pointwise convergence theorem, the Fourier series of the function f (x) converges to the sum cosx if π< x <, S(x) =, если x =, x = ±π, cosx, если < x < π. В результате функция f(x) = cosx, заданная на промежутке (, π), выражена через синусы: cosx = 8 π k=1 k 4k 2 1 sin 2kx, x (, π). Рис демонстрируют постепенное приближение частичных сумм S 1 (x), S 2 (x), S 3 (x) к разрывной функции f (x). 26

27 Fig. Fig. 18. Graph of the function f (x) with the graph of the partial sum S 1 (x) superimposed on it. 19. Graph of the function f(x) with the graph of the partial sum S 2 (x) superimposed on it 27

28 Fig. Fig. 2. Graph of the function f (x) with the graph of the partial sum S 3 (x) superimposed on it. 21 shows graphs of the function f (x) and its partial sum S 99 (x). Rice. 21. Graph of the function f (x) with a graph of the partial sum S 99 (x) 28 superimposed on it

29 EXAMPLE 8. Let us expand the function f(x) = e ax, a >, x [, π], in a Fourier series only in cosines. Solution. We continue the function in an even way to (,) (i.e., so that the equality f(x) = f(x) holds for all x (, π)), and then periodically with a period of 2π to the entire real axis. We obtain the function f (x), the graph of which is shown in Fig. 22. Function f (x) at points 22. The graph of the continued function f (x) x = kπ, k is an integer, has kinks. Let us calculate the Fourier coefficients: b n =, since f (x) is even. Integrating by parts, we get 29

30 a n = 2 π a = 2 π = 2 cosnxd(e ax) = 2 πa e ax dx = 2 π a (eaπ 1), f(x) cos πnxdx = 2 π πa eax cosnx = 2 πa (eaπ cosnπ 1 ) + 2n πa 2 π e ax cos nxdx = + 2n e ax sin nxdx = πa sin nxde ax = = 2 π a (eaπ cos n π 1) + 2n π sin nx π a 2eax 2n2 e ax cos nxdx = 2 π a 2 π a (eaπ cos n π 1) n2 a a n. 2 Therefore, a n = 2a e aπ cos n π 1. π a 2 + n 2 Since f (x) is continuous, according to the pointwise convergence theorem, its Fourier series converges to f (x). Hence, for all x [, π] we have f(x) = 1 π a (eaπ 1)+ 2a π k=1 e aπ (1) k 1 a 2 + k 2 coskx (x π). Figures demonstrate the gradual approximation of the partial sums of the Fourier series to a given discontinuous function. 3

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

32 Fig. 27. Graphs of functions f (x) and S 4 (x) 28. Graphs of the functions f (x) and S 99 (x) PROBLEM 9. Expand the function f (x) = cos x, x π, in a Fourier series only in cosines. 1. Expand the function f (x) \u003d e ax, a >, x π, in a Fourier series only in terms of sines. 11. Expand the function f (x) \u003d x 2, x π, in a Fourier series only in sines. 12. Expand the function f (x) \u003d sin ax, x π, in a Fourier series in terms of cosines only. 13. Expand the function f (x) \u003d x sin x, x π, in a Fourier series only in sines. Answers 9. cosx = cosx. 1. e ax = 2 [ 1 (1) k e aπ] k sin kx. π a 2 + k2 k=1 11. x 2 2 [ π 2 (1) n 1 π n + 2 ] n 3 ((1)n 1) sin nx. 32

33 12. If a is not an integer, then sin ax = 1 cosaπ (1 + +2a cos 2nx ) + π a 2 (2n) 2 +2a 1 + cosaπ cos(2n 1)x π a 2 (2n 1) 2; if a = 2m is an even number, then sin 2mx = 8m cos(2n 1)x π (2m) 2 (2n 1) 2; if a = 2m 1 is a positive odd number, then sin(2m 1)x = 2 ( cos 2nx ) 1 + 2(2m 1). π (2m 1) 2 (2n) π 16 n sin x sin 2nx. 2 π (4n 2 1) 2 3. Fourier series of a function with an arbitrary period Assume that the function f(x) is defined in the interval [ l, l], l >. By substituting x = ly, y π, we obtain the function g(y) = f(ly/π) defined in the interval π [, π]. This function g(y) corresponds to the (formal) Fourier series () ly f = g(y) a π 2 + (a n cosny + b n sin ny), whose coefficients are found by the Euler Fourier formulas: a n = 1 π g(y) cosny dy = 1 π f (ly π) cos ny dy, n =, 1, 2,..., 33

34 b n = 1 π g(y) sinny dy = 1 π f () ly sin ny dy, n = 1, 2,.... π l, we obtain a slightly modified trigonometric series for the function f(x): where f(x) a 2 + a n = 1 l b n = 1 l l l l l (a n cos πnx l f(x) cos πnx l f(x) sin πnx l + b n sin πnx), (11) l dx, n =, 1, 2,..., (12) dx, n = 1, 2,.... (13) Formulas (11) (13) are said to define expansion in a Fourier series of a function with an arbitrary period. EXAMPLE 9. Find the Fourier series of the function given in the interval (l, l) by the expression ( A if l< x, f(x) = B, если < x < l, считая, что она периодична с периодом 2l. Решение. Продолжим функцию периодически, с периодом 2l, на всю ось. Получим функцию f (x), кусочно-постоянную в промежутках (l + 2kl, l + 2kl), и претерпевающую разрывы первого рода в точках x = lk, k целое число. Ее коэффициенты Фурье вычисляются по формулам (12) и (13): 34

35 a = 1 l l f(x) dx = 1 l A dx + 1 l l B dx = A + B, l l a n = 1 l l l f(x) cos πnx l dx = = 1 l = 1 l l A cos πnx l = A + B π n l b n = 1 l dx + 1 l l B cos πnx l sin πn = if n, l l A sin πnx l f(x) sin πnx l dx + 1 l l dx = B sin πnx l = B A (1 cosπn). πn Compose the Fourier series of the function f (x) : f(x) A + B π (B A Since cosπn = (1) n, then n dx = dx = (1 cosπn) sin πnx). l for n = 2k we get b n = b 2k =, for n = 2k 1 b n = b 2k 1 = 35 2(B A) π(2k 1).

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

37 Fig. 29. Graph of the function f (x) with superimposed graphs of the harmonics S (x) = a 2 and S 1 (x) = b 1 sinx. For clarity, the graphs of the three higher harmonics S 3 (x) \u003d b 3 sin 3πx, S l 5 (x) \u003d b 5 sin 5πx l and S 7 (x) \u003d b 7 sin 7πx are shifted vertically up l 37

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

39 PROBLEMS In problems, expand the specified functions in Fourier series in given intervals. 14. f(x) = x 1, (1, 1). 15. f(x) = ch2x, (2, 2] f(x) = x (1 x), (1, 1]. 17. f(x) = cos π x, [ 1, 1] f(x ) = sin π x, (1, 1).( 2 1 if 1< x < 1, 19. f(x) = 2l = 4., если 1 < x < 3; x, если x 1, 2. f(x) = 1, если 1 < x < 2, 2l = 3. { 3 x, если 2 x < 3;, если ωx, 21. f(x) = 2l = 2π/ω. sin ωx, если ωx π; Разложить в ряды Фурье: а) только по косинусам; б) только по синусам указанные функции в заданных промежутках (, l) { 22. f(x) = { 23. f(x) = ax, если < x < l/2, a(l x), если l/2 < x < l. 1, если < x 1, 2 x, если 1 x 2. Ответы 14. f(x) = 4 cos(2n 1)πx. π 2 (2n 1) f(x) = sh sh4 (1) n nπx cos 16 + π 2 n f(x) = cos 2nπx. π 2 n f(x) = 2 π + 8 π (1) n n 1 4n 2 cosnπx. 39

40 18. f(x) = 8 (1) n n sin nπx. π 1 4n (1) n 2n + 1 cos πx. π 2n πn 2πnx π 2 sin2 cos n π sin ωx 2 cos 2nωx π 4n 2 1. (l 22. a) f(x) = al 4 2) 1 (4n 2)πx cos, π 2 (2n 1) 2 l b) f(x) = 4al (1) n 1 (2n 1) πx sin. π 2 (2n 1) 2 l 23. a) f(x) = (cos π π 2 2 x 2 2 cos 2π 2 2 x cos 3π 2 2 x cos 5π), 2 2 x... b) f( x) = 4 (sin π π 2 2 x 1 3 sin 3π)+ 2 2 x (sin π π 2 x cos 2π) 2 x Complex form of the Fourier series Decomposition f(x) = c n e inx, where c n = 1 2π f (x)e inx dx, n = ±1, ±2,..., is called the complex form of the Fourier series. The function expands into a complex Fourier series under the same conditions under which it expands into a real Fourier series. four

41 EXAMPLE 1. Find the Fourier series in the complex form of the function given by the formula f(x) = e ax in the interval [, π), where a is a real number. Solution. Let us calculate the coefficients: = c n = 1 2π f(x)e inx dx = 1 2π e (a in)x dx = 1 ((1) n e aπ (1) n e aπ) = (1)n sh aπ. 2π(a in) π(a in) The complex Fourier series of the function f has the form f(x) sh aπ π n= (1) n a in einx. Let us verify that the function f(x) is piecewise smooth: in the interval (, π) it is continuously differentiable, and at the points x = ±π there are finite limits (5), (6) lim h + ea(+h) = e aπ, lim h + ea(π h) = e aπ, e a(+h) e a(+) lim h + h = ae aπ e a(π h) e a(π), lim h + h = ae aπ. Therefore, the function f(x) can be represented by a Fourier series sh aπ π n= (1) n a in einx, which converges to the sum: ( e S(x) = ax if π< x < π, ch a, если x = ±π. 41

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

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

44 Now let's find the Fourier series in real form. To do this, we group the terms with numbers n and n for n: a n e inx + a n e inx = 2a neinx + e inx Since c = 1, then 2 = 2a n cos nx. f(x) = 1 a 2 1 2a cosx + a = a n cosnx. 2 This is a Fourier series in the real form of the function f(x). Thus, without calculating a single integral, we found the Fourier series of the function. In doing so, we calculated a hard integral depending on the parameter cos nxdx 1 2a cosx + a = 2 π an 2 1 a2, a< 1. (15) ПРИМЕР 12. Найдем ряд Фурье в комплексной и вещественной форме функции, заданной формулой a sin x f(x) = 1 2a cosx + a2, a < 1, a R. Решение. Функция f(x) является нечетной, поэтому для всех n a n = и b n = 2 π f(x) sin nxdx = 2a π sin x sin nxdx 1 2a cosx + a 2. Чтобы записать ряд Фурье нужно вычислить сложные интегралы или воспользоваться приемом, описанным выше. Поступим вторым способом: 44

45 a(z z 1) f(x) = 2i (1 a(z z 1) + a 2) = i 2 + i (a + a 1)z 2 2 (z a)(z a 1) = = i 2 + i () a 2 z a + a 1. z a 1 We expand each of the simple fractions according to the geometric progression formula: + a z a = a 1 z 1 a = a a n z z n, n= z a 1 z a = az = a n z n. n= This is possible because az = a/z = a< 1. Значит + ia n /2, если n <, f(x) c n e inx, где c n =, если n =, n= ia n /2, если n >, or, more briefly, c n = 1 2i a n sgnn. Thus, the Fourier series in complex form is found. Grouping terms with numbers n and n, we obtain the Fourier series of the function in real form: = f(x) = + a sin x 1 2a cosx + a + 2 (1 2i an e inx 1 2i an e inx n= +) = c n e inx = a n sin nx. Again, we managed to calculate the following complex integral: sin x sin nxdx 1 2a cosx + a 2 = π an 1. (16) 45

46 PROBLEM 24. Using (15), calculate the integral cos nxdx 1 2a cosx + a 2 for real a, a > Using (16), calculate the integral sin x sin nxdx for real a, a > a cosx + a2 In problems, find the series Fourier in complex form for functions. 26. f(x) = sgn x, π< x < π. 27. f(x) = ln(1 2a cosx + a 2), a < 1. 1 a cosx 28. f(x) = 1 2a cosx + a2, a < Докажите, что функция f, определенная в промежутке [, π], вещественнозначна, если и только если коэффициенты c n ее комплексного ряда Фурье связаны соотношениями c n = c n, n =, ±1, ±2, Докажите, что функция f, определенная в промежутке [, π], является четной (т. е. удовлетворяет соотношению f(x) = f(x)), если и только если коэффициенты c n ее комплексного ряда Фурье связаны соотношениями c n = c n, n = ±1, ±2, Докажите, что функция f, определенная в промежутке [, π], является нечетной (т. е. удовлетворяет соотношению f(x) = f(x)), если и только если коэффициенты c n ее комплексного ряда Фурье связаны соотношениями c n = c n, n =, ±1, ±2,.... Ответы 1 2π 24. a n a π a n i + e 2inx, где подразумевается, что слагаемое, соответствующее n =, пропущено. π n n= a n n cosnx. 28. a n cosnx. n= 46

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

48 a n = 2 π f(x) cosnxdx = 2 π a cos nxdx = 2 sin na πn. Therefore, the Lyapunov equality for the function f(x) takes the form: 2 a 2 π + 4 sin 2 na = 2a 2 π 2 n 2 π. From the last equality for a π we find sin 2 na n 2 = a(π a) 2 Assuming a = π 2, we obtain sin2 na = 1 for n = 2k 1 and sin 2 na = for n = 2k. Therefore, k=1 1 (2k 1) 2 = π2 8. EXAMPLE 14. Let's write the Lyapunov equality for the function f(x) = x cosx, x [, π], and use it to find the sum of the number series (4n 2 + 1) 2 (4n 2 1) 4. 1 π Solution. Direct calculations give = π π f 2 (x) dx = 1 π x 2 cos 2 xdx = 1 π x sin 2xdx = π π x cos x = π x 21 + cos 2x dx = 2 π 1 4π cos 2xdx =

49 Since f(x) is an even function, then for all n we have b n =, a n = 2 π = 1 π 1 = π(n + 1) = f(x) cosnxdx = 2 π 1 cos(n + 1)x π (n + 1) 2 x cosxcosnxdx = x (cos(n + 1)x + cos(n 1)x) dx = 1 π sin(n + 1)xdx sin(n 1)xdx = π(n 1) π π 1 + cos(n 1)x = π(n 1) 2 1 (= (1) (n+1) 1) 1 (+ (1) (n+1) 1) = π(n + 1) 2 π(n 1) 2 () = (1)(n+1) 1 1 π (n + 1) + 1 = 2 (n 1) 2 = 2 (1)(n+1) 1 n k π (n 2 1) = π (4k 2 1) 2 if n = 2k, 2 if n = 2k + 1. The coefficient a 1 must be calculated separately, since in the general formula for n = 1 the denominator of the fraction vanishes. = 1 π a 1 = 2 π f(x) cosxdx = 2 π x(1 + cos 2x)dx = π 2 1 2π 49 x cos 2 xdx = sin 2xdx = π 2.

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

51 π (2n + 1) = π sin 2απ 2απ = 2sin2 απ α 2 π 2 Answers + 4 sin2 απ π 2 α 2 (α 2 n 2) 2; sin 2απ 1 2απ = απ n 2 4sin2 π 2 (α 2 n 2) 2. 1 π 35. f(x)g(x) dx= c n d n, where c n is the Fourier coefficient 2π of f(x), and d n is the Fourier coefficient functions g(x). 6. Differentiation of Fourier series Let f: R R be a continuously differentiable 2π-periodic function. Its Fourier series has the form: f(x) = a 2 + (a n cos nx + b n sin nx). The derivative f (x) of this function will be a continuous and 2π-periodic function, for which a formal Fourier series can be written: f (x) a 2 + (a n cos nx + b n sin nx), where a, a n, b n, n = 1 , 2,... Fourier coefficients of the function f (x). 51

52 Theorem (on term-by-term differentiation of Fourier series). Under the above assumptions, the equalities a =, a n = nb n, b n = na n, n 1 are true. EXAMPLE 15. Let a piecewise-smooth function f(x) be continuous in the interval [, π]. Let us prove that when the condition f(x)dx = is satisfied, the inequality 2 dx 2 dx, called the Steklov inequality, holds, and we verify that equality in it is realized only for functions of the form f(x) = A cosx. In other words, Steklov's inequality gives conditions under which the smallness of the derivative (in the mean square) implies the smallness of the function (in the mean square). Solution. Let us extend the function f(x) to the interval [, ] evenly. Denote the extended function by the same symbol f(x). Then the continued function will be continuous and piecewise smooth on the interval [, π]. Since the function f(x) is continuous, then f 2 (x) is continuous on the interval and 2 dx< +, следовательно, можно применить теорему Ляпунова, согласно которой имеет место равенство 1 π 2 dx = a () a 2 n + b 2 n. 52

53 Since the continued function is even, then b n =, a = by condition. Consequently, the Lyapunov equality takes the form 1 π 2 dx = a 2 π n. (17) Let us make sure that f (x) satisfies the conclusion of the theorem on term-by-term differentiation of the Fourier series, that is, that a =, a n = nb n, b n = na n, n 1. Let the derivative f (x) undergo breaks at the points x 1, x 2,..., x N in the interval [, π]. Denote x =, x N+1 = π. Let us divide the integration interval [, π] into N +1 intervals (x, x 1),..., (x N, x N+1), on each of which f(x) is continuously differentiable. Then, using the additivity property of the integral and then integrating by parts, we get: b n = 1 π = 1 π = 1 π f (x) sin nxdx = 1 π N f(x) sin nx j= N f(x) sin nx j= x j+1 x j x j+1 x j n n π N j= x j+1 x j x j+1 x j f (x) sin nxdx = f(x) cosnxdx = f(x) cosnxdx = = 1 π [(f(x 1) sin nx 1 f(x) sin nx) + + (f(x 2) sinnx 2 f(x 1) sin nx 1)

54 + (f(x N+1) sin nx N+1 f(x N) sin nx N)] na n = = 1 π na n = = 1 π na n = na n. x j+1 a = 1 f (x)dx = 1 N f (x)dx = π π j= x j = 1 N x j+1 f(x) π = 1 (f(π) f()) = . x j π j= Similarly, we get a n = nb n. We have shown that the theorem on term-by-term differentiation of Fourier series for a continuous piecewise-smooth 2π-periodic function whose derivative in the interval [, π] undergoes discontinuities of the first kind is true. So f (x) a 2 + (a n cosnx + b n sin nx) = (na n)sin nx, since a =, a n = nb n =, b n = na n, n = 1, 2,.... Because 2dx< +, то по равенству Ляпунова 1 π 2 dx = 54 n 2 a 2 n. (18)

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

56 PROBLEMS 37. Let a piecewise-smooth function f(x) be continuous on the interval [, π]. Prove that under the condition f() = f(π) = the inequality 2 dx 2 dx, also called Steklov's inequality, holds, and make sure that equality in it holds only for functions of the form f(x) = B sin x. 38. Let a function f be continuous in the interval [, π] and have in it (with the possible exception of only a finite number of points) a square-integrable derivative f (x). Prove that if, in addition, the conditions f() = f(π) and f(x) dx = are satisfied, then the inequality 2 dx 2 dx, called the Wirtinger inequality, holds, and the equality in it takes place only for functions of the form f(x ) = A cosx + B sinx. 56

57 7. Application of Fourier series for solving partial differential equations When studying a real object (natural phenomena, production process, control system, etc.), two factors turn out to be significant: the level of accumulated knowledge about the object under study and the degree of development of the mathematical apparatus. At the present stage of scientific research, the following chain has been developed: a phenomenon a physical model a mathematical model. The physical formulation (model) of the problem is as follows: the conditions for the development of the process and the main factors influencing it are identified. The mathematical formulation (model) consists in describing the factors and conditions chosen in the physical formulation in the form of a system of equations (algebraic, differential, integral, etc.). A problem is said to be well-posed if, in a certain functional space, the solution of the problem exists, uniquely and continuously depends on the initial and boundary conditions. The mathematical model is not identical to the object under consideration, but is its approximate description Derivation of the equation of free small transverse vibrations of the string We will follow the textbook. Let the ends of the string be fixed, and the string itself be taut. If the string is taken out of equilibrium (for example, by pulling or striking it), then the string will start 57

58 hesitate. We will assume that all points of the string move perpendicular to its equilibrium position (transverse vibrations), and at each moment of time the string lies in the same plane. Let us take a system of rectangular coordinates xou in this plane. Then, if at the initial time t = the string was located along the axis Ox, then u will mean the deviation of the string from the equilibrium position, that is, the position of the string point with the abscissa x at an arbitrary time t corresponds to the value of the function u(x, t). For each fixed value of t, the graph of the function u(x, t) represents the shape of the vibrating string at time t (Fig. 32). At a constant value of x, the function u(x, t) gives the law of motion of a point with the abscissa x along a straight line parallel to the Ou axis, the derivative u t is the speed of this motion, and the second derivative 2 u t 2 is the acceleration. Rice. 32. Forces applied to an infinitely small section of a string Let's write an equation that the function u(x, t) must satisfy. To do this, we make some more simplifying assumptions. We will assume that the string is absolutely flexible.

59 coy, that is, we will assume that the string does not resist bending; this means that the stresses arising in the string are always directed tangentially to its instantaneous profile. The string is assumed to be elastic and subject to Hooke's law; this means that the change in the magnitude of the tension force is proportional to the change in the length of the string. Let us assume that the string is homogeneous; this means that its linear density ρ is constant. We neglect external forces. This means that we are considering free oscillations. We will study only small vibrations of a string. If we denote by ϕ(x, t) the angle between the abscissa axis and the tangent to the string at the point with the abscissa x at time t, then the condition for small oscillations is that the value of ϕ 2 (x, t) can be neglected in comparison with ϕ (x, t), i.e., ϕ 2. Since the angle ϕ is small, then cos ϕ 1, ϕ sin ϕ tg ϕ u, therefore, the value (u x x,) 2 can also be neglected. It immediately follows from this that in the process of oscillation we can neglect the change in the length of any section of the string. Indeed, the length of a piece of string M 1 M 2 projected into the interval of the x-axis, where x 2 = x 1 + x, is equal to l = x 2 x () 2 u dx x. x Let us show that, under our assumptions, the value of the tension force T will be constant along the entire string. To do this, we take some part of the string M 1 M 2 (Fig. 32) at time t and replace the action of the discarded parts

60 kov by the tension forces T 1 and T 2. Since, according to the condition, all points of the string move parallel to the Ou axis and there are no external forces, the sum of the projections of the tension forces on the Ox axis must be equal to zero: T 1 cosϕ(x 1, t) + T 2 cosϕ(x 2, t) =. Hence, due to the smallness of the angles ϕ 1 = ϕ(x 1, t) and ϕ 2 = ϕ(x 2, t), we conclude that T 1 = T 2. Denote the general value of T 1 = T 2 by T. Now we calculate the sum of projections F u of the same forces on the Ou axis: F u = T sin ϕ(x 2, t) T sin ϕ(x 1, t). (2) Since for small angles sin ϕ(x, t) tg ϕ(x, t), and tg ϕ(x, t) u(x, t)/ x, equation (2) can be rewritten as F u T (tan ϕ(x 2, t) tan ϕ(x 1, t)) (u T x (x 2, t) u) x (x 1, t) x x T 2 u x 2(x 1, t) x . Since the point x 1 is chosen arbitrarily, then F u T 2 u x2(x, t) x. After all the forces acting on the section M 1 M 2 are found, we apply Newton's second law to it, according to which the product of mass and acceleration is equal to the sum of all acting forces. The mass of a piece of string M 1 M 2 is equal to m = ρ l ρ x, and the acceleration is equal to 2 u(x, t). Newton's t 2 equation takes the form: 2 u t (x, t) x = u 2 α2 2 x2(x, t) x, where α 2 = T ρ is a constant positive number. 6

61 Reducing by x, we get 2 u t (x, t) = u 2 α2 2 x2(x, t). (21) As a result, we have obtained a linear homogeneous partial differential equation of the second order with constant coefficients. It is called the string vibration equation or the one-dimensional wave equation. Equation (21) is essentially a reformulation of Newton's law and describes the motion of a string. But in the physical formulation of the problem, there were requirements that the ends of the string are fixed and the position of the string at some point in time is known. We will write these conditions in equations as follows: a) we will assume that the ends of the string are fixed at the points x = and x = l, i.e., we will assume that for all t the relations u(, t) =, u(l, t ) = ; (22) b) we will assume that at the time t = the position of the string coincides with the graph of the function f(x), i.e., we will assume that for all x [, l] the equality u(x,) = f( x); (23) c) we will assume that at the time t = the point of the string with the abscissa x is given speed g(x), i.e., we will assume that u (x,) = g(x). (24) t Relations (22) are called boundary conditions, and relations (23) and (24) are called initial conditions. Mathematical model of free small transverse 61

62 string vibrations is that it is necessary to solve equation (21) with boundary conditions (22) and initial conditions (23) and (24) Solution of the equation of free small transverse vibrations of the string by the Fourier method< t <, удовлетворяющие граничным условиям (22) и начальным условиям (23) и (24), будем искать методом Фурье (называемым также методом разделения переменных). Метод Фурье состоит в том, что частные решения ищутся в виде произведения двух функций, одна из которых зависит только от x, а другая только от t. То есть мы ищем решения уравнения (21), которые имеют специальный вид: u(x, t) = X(x)T(t), (25) где X дважды непрерывно дифференцируемая функция от x на [, l], а T дважды непрерывно дифференцируемая функция от t, t >. Substituting (25) into (21), we get: X T = α 2 X T, (26) or T (t) α 2 T(t) = X (x) X(x). (27) It is said that there has been a separation of variables. Since x and t do not depend on each other, the left side in (27) does not depend on x, but the right side does not depend on t, and the total value of these ratios is 62

63 must be constant, which we denote by λ: T (t) α 2 T(t) = X (x) X(x) = λ. Hence we obtain two ordinary differential equations: X (x) λx(x) =, (28) T (t) α 2 λt(t) =. (29) In this case, the boundary conditions (22) take the form X()T(t) = and X(l)T(t) =. Since they must be fulfilled for all t, t >, then X() = X(l) =. (3) Let us find solutions to equation (28) satisfying boundary conditions (3). Let's consider three cases. Case 1: λ >. Denote λ = β 2. Equation (28) takes the form X (x) β 2 X(x) =. Its characteristic equation k 2 β 2 = has roots k = ±β. Therefore, the general solution of Eq. (28) has the form X(x) = C e βx + De βx. We must choose the constants C and D so that the boundary conditions (3) are met, i.e. X() = C + D =, X(l) = C e βl + De βl =. Since β, then this system of equations has a unique solution C = D =. Hence X(x) and 63

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

65 In what follows, we will assign to n only positive values ​​n = 1, 2,..., since for negative n, solutions of the same form (nπ) will be obtained. The values ​​λ n = are called eigenvalues, and the functions X n (x) = C n sin πnx eigenfunctions of differential equation (28) with boundary conditions (3). Now let's solve equation (29). For him, the characteristic equation has the form k 2 α 2 λ =. (32) l 2 Since we found out above that nontrivial solutions X(x) of Eq. (28) exist only for negative λ equal to λ = n2 π 2, it is these λ that we will consider below. The roots of equation (32) are k = ±iα λ, and the solutions of equation (29) have the form: T n (t) = A n sin πnαt + B n cos πnαt, (33) l l where A n and B n are arbitrary constants. Substituting formulas (31) and (33) into (25), we find particular solutions of equation (21) that satisfy boundary conditions (22): (u n (x, t) = B n cos πnαt + A n sin πnαt) C n sin pnx. l l l Entering the factor C n in brackets and introducing the notation C n A n = b n and B n C n = a n, we write u n (X, T) as (u n (x, t) = a n cos πnαt + b n sin πnαt) sin pnx. (34) l l l 65

66 The vibrations of the string corresponding to the solutions u n (x, t) are called natural vibrations of the string. Since equation (21) and boundary conditions (22) are linear and homogeneous, then a linear combination of solutions (34) (u(x, t) = a n cos πnαt + b n sin πnαt) sin πnx (35) l l l will be a solution to equation (21 ) satisfying the boundary conditions (22) with a special choice of the coefficients a n and b n, which ensures the uniform convergence of the series. Now we select the coefficients a n and b n of solution (35) so that it satisfies not only the boundary conditions, but also the initial conditions (23) and (24), where f(x), g(x) are given functions (moreover, f() = f (l) = g() = g(l) =). We assume that the functions f(x) and g(x) satisfy the Fourier expansion conditions. Substituting the value t = into (35), we obtain u(x,) = a n sin πnx l = f(x). Differentiating series (35) with respect to t and substituting t =, we obtain u t (x,) = πnα b n sin πnx l l = g(x), and this is the expansion of the functions f(x) and g(x) into Fourier series. Therefore, a n = 2 l l f(x) sin πnx l dx, b n = 2 l g(x) sin πnx dx. πnα l (36) 66

67 Substituting the expressions for the coefficients a n and b n into series (35), we obtain a solution to equation (21) that satisfies boundary conditions (22) and initial conditions (23) and (24). Thus, we have solved the problem of free small transverse vibrations of a string. Let us clarify the physical meaning of the eigenfunctions u n (x, t) of the problem of free vibrations of a string, defined by formula (34). Let us rewrite it as where u n (x, t) = α n cos πnα l α n = a 2 n + b2 n, (t + δ n) sin πnx, (37) l πnα δ n = arctg b n. l a n Formula (37) shows that all points of the string perform harmonic oscillations with the same frequency ω n = πnα and phase πnα δ n. The oscillation amplitude depends on l l the abscissa x of the string point and is equal to α n sin πnx. With such an oscillation, all points of the string simultaneously reach their l maximum deviation in one direction or another and simultaneously pass the equilibrium position. Such oscillations are called standing waves. A standing wave will have n + 1 fixed points given by the roots of the equation sin πnx = in the interval [, l]. The fixed points are called the nodes of the standing wave. In the middle between the nodes - l mi are the points at which the deviations reach a maximum; such points are called antinodes. Each string can have its own oscillations of strictly defined frequencies ω n = πnα, n = 1, 2,.... These frequencies are called natural frequencies of the string. The lowest l tone that a string can produce is determined by itself 67

68 low natural frequency ω 1 = π T and is called the fundamental tone of the string. The remaining tones corresponding to l ρ frequencies ω n, n = 2, 3,..., are called overtones or harmonics. For clarity, we will depict the typical profiles of a string emitting the fundamental tone (Fig. 33), the first overtone (Fig. 34) and the second overtone (Fig. 35). Rice. Fig. 33. Profile of the string that emits the fundamental tone. Fig. 34. Profile of a string emitting the first overtone. Fig. 35. Profile of a string emitting a second overtone If the string performs free vibrations determined by the initial conditions, then the function u(x, t) is represented, as can be seen from formula (35), as a sum of individual harmonics. Thus arbitrary oscillation 68

The 69th string is a superposition of standing waves. In this case, the nature of the sound of the string (tone, sound strength, timbre) will depend on the ratio between the amplitudes of individual harmonics. Strength, pitch and timbre of the sound A vibrating string excites air vibrations perceived by the human ear as a sound emitted by a string. The strength of sound is characterized by the energy or amplitude of vibrations: the greater the energy, the greater the strength of the sound. The pitch of a sound is determined by its frequency or period of oscillation: the higher the frequency, the higher the sound. The timbre of sound is determined by the presence of overtones, the distribution of energy over harmonics, i.e., the method of excitation of oscillations. The amplitudes of the overtones are, generally speaking, less than the amplitude of the fundamental, and the phases of the overtones can be arbitrary. Our ear is not sensitive to the phase of oscillations. Compare, for example, the two curves in Fig. 36, borrowed from . This is a recording of sound with the same fundamental tone, extracted from the clarinet (a) and the piano (b). Both sounds are not simple sinusoidal oscillations. The fundamental frequency of the sound in both cases is the same and this creates the same tone. But the curve patterns are different because different overtones are superimposed on the fundamental tone. In a sense, these drawings show what timbre is. 69

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Fourier series of periodic functions with period 2π.

The Fourier series allows you to study periodic functions by decomposing them into components. Alternating currents and voltages, displacements, speed and acceleration of crank mechanisms, and acoustic waves are typical practical applications of periodic functions in engineering calculations.

The Fourier series expansion is based on the assumption that all functions of practical importance in the interval -π ≤ x ≤ π can be expressed as convergent trigonometric series (a series is considered convergent if the sequence of partial sums made up of its terms converges):

Standard (=usual) notation through the sum of sinx and cosx

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

where a o , a 1 ,a 2 ,...,b 1 ,b 2 ,.. are real constants, i.e.

Where, for the range from -π to π, the coefficients of the Fourier series are calculated by the formulas:

The coefficients a o ,a n and b n are called Fourier coefficients, and if they can be found, then series (1) is called near Fourier, corresponding to the function f(x). For series (1), the term (a 1 cosx+b 1 sinx) is called the first or main harmonica,

Another way to write a series is to use the relation acosx+bsinx=csin(x+α)

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

Where a o is a constant, c 1 \u003d (a 1 2 +b 1 2) 1/2, c n \u003d (a n 2 +b n 2) 1/2 are the amplitudes of the various components, and is equal to a n \u003d arctg a n /b n.

For series (1), the term (a 1 cosx + b 1 sinx) or c 1 sin (x + α 1) is called the first or main harmonica,(a 2 cos2x+b 2 sin2x) or c 2 sin(2x+α 2) is called second harmonic and so on.

To accurately represent a complex signal, an infinite number of terms is usually required. However, in many practical problems it is sufficient to consider only the first few terms.

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

Expansion of non-periodic functions in a Fourier series.

If the function f(x) is non-periodic, then it cannot be expanded in a Fourier series for all values ​​of x. However, it is possible to define a Fourier series representing a function over any range of width 2π.

Given a non-periodic function, one can compose a new function by choosing f(x) values ​​within a certain range and repeating them outside this range at 2π intervals. Since the new function is periodic with a period of 2π, it can be expanded in a Fourier series for all values ​​of x. For example, the function f(x)=x is not periodic. However, if it is necessary to expand it into a Fourier series on the interval from 0 to 2π, then a periodic function with a period of 2π is constructed outside this interval (as shown in the figure below).

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

Even and odd functions.

They say the function y=f(x) even if f(-x)=f(x) for all values ​​of x. Graphs of even functions are always symmetrical about the y-axis (that is, they are mirrored). Two examples of even functions: y=x 2 and y=cosx.

They say that the function y=f(x) odd, if f(-x)=-f(x) for all values ​​of x. Graphs of odd functions are always symmetrical about the origin.

Many functions are neither even nor odd.

Fourier series expansion in cosines.

The Fourier series of an even periodic function f(x) with period 2π contains only cosine terms (i.e., does not contain sine terms) and may include a constant term. Consequently,

where are the coefficients of the Fourier series,

The Fourier series of an odd periodic function f(x) with period 2π contains only terms with sines (i.e., does not contain terms with cosines).

Consequently,

where are the coefficients of the Fourier series,

Fourier series on a half-cycle.

If a function is defined for a range, say 0 to π, and not just 0 to 2π, it can be expanded into a series only in terms of sines or only in terms of cosines. The resulting Fourier series is called near Fourier on a half cycle.

If you want to get a decomposition Fourier on a half-cycle in cosines functions f(x) in the range from 0 to π, then it is necessary to compose an even periodic function. On fig. below is the function f(x)=x built on the interval from x=0 to x=π. Since the even function is symmetrical about the f(x) axis, we draw the line AB, as shown in Fig. below. If we assume that outside the considered interval, the resulting triangular shape is periodic with a period of 2π, then the final graph has the form, display. in fig. below. Since it is required to obtain the Fourier expansion in cosines, as before, we calculate the Fourier coefficients a o and a n

If you want to get functions f (x) in the range from 0 to π, then you need to compose an odd periodic function. On fig. below is the function f(x)=x built on the interval from x=0 to x=π. Since the odd function is symmetric with respect to the origin, we construct the line CD, as shown in Fig. If we assume that outside the considered interval, the received sawtooth signal is periodic with a period of 2π, then the final graph has the form shown in Fig. Since it is required to obtain the Fourier expansion on a half-cycle in terms of sines, as before, we calculate the Fourier coefficient. b

Fourier series for an arbitrary interval.

Expansion of a periodic function with period L.

The periodic function f(x) repeats as x increases by L, i.e. f(x+L)=f(x). The transition from the previously considered functions with period 2π to functions with period L is quite simple, since it can be done using a change of variable.

To find the Fourier series of the function f(x) in the range -L/2≤x≤L/2, we introduce a new variable u so that the function f(x) has a period of 2π with respect to u. If u=2πx/L, then x=-L/2 for u=-π and x=L/2 for u=π. Also let f(x)=f(Lu/2π)=F(u). The Fourier series F(u) has the form

Where are the coefficients of the Fourier series,

More often, however, the above formula leads to dependence on x. Since u=2πх/L, then du=(2π/L)dx, and the limits of integration are from -L/2 to L/2 instead of -π to π. Therefore, the Fourier series for the dependence on x has the form

where in the range from -L/2 to L/2 are the coefficients of the Fourier series,

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

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

For the substitution u=πx/L, the interval from x=0 to x=L corresponds to the interval from u=0 to u=π. Therefore, the function can be expanded into a series only in terms of cosines or only in terms of sines, i.e. in Fourier series on a half cycle.

The expansion in cosines in the range from 0 to L has the form