Fourier series expansion according to the graph examples. Expansion of functions into power series

How to insert mathematical formulas on the site?

If you ever need to add one or two mathematical formulas to a web page, then the easiest way to do this is as described in the article: mathematical formulas are easily inserted into the site in the form of pictures that Wolfram Alpha automatically generates. In addition to simplicity, this universal method will help improve the visibility of the site in search engines. It has been working for a long time (and I think it will work forever), but it is morally outdated.

If you are constantly using math formulas on your site, then I recommend you use MathJax, a custom JavaScript library that displays math notation in web browsers using MathML, LaTeX, or ASCIIMathML markup.

There are two ways to start using MathJax: (1) using a simple code, you can quickly connect a MathJax script to your site, which will be automatically loaded from a remote server at the right time (list of servers); (2) upload the MathJax script from a remote server to your server and connect it to all pages of your site. The second method is more complicated and time consuming and will allow you to speed up the loading of the pages of your site, and if the parent MathJax server becomes temporarily unavailable for some reason, this will not affect your own site in any way. Despite these advantages, I chose the first method, as it is simpler, faster and does not require technical skills. Follow my example, and within 5 minutes you will be able to use all the features of MathJax on your site.

You can connect the MathJax library script from a remote server using two code options taken from the main MathJax website or from the documentation page:

One of these code options needs to be copied and pasted into your web page code, preferably between tags and or right after the tag . According to the first option, MathJax loads faster and slows down the page less. But the second option automatically tracks and loads the latest versions of MathJax. If you insert the first code, then it will need to be updated periodically. If you paste the second code, then the pages will load more slowly, but you will not need to constantly monitor MathJax updates.

The easiest way to connect MathJax is in Blogger or WordPress: in the site control panel, add a widget designed to insert third-party JavaScript code, copy the first or second version of the load code presented above into it, and place the widget closer to the beginning of the template (by the way, this is not necessary at all , since the MathJax script is loaded asynchronously). That's all. Now learn the MathML, LaTeX, and ASCIIMathML markup syntax and you're ready to embed math formulas into your web pages.

Any fractal is built according to a certain rule, which is consistently applied an unlimited number of times. Each such time is called an iteration.

The iterative algorithm for constructing a Menger sponge is quite simple: the original cube with side 1 is divided by planes parallel to its faces into 27 equal cubes. One central cube and 6 cubes adjacent to it along the faces are removed from it. It turns out a set consisting of 20 remaining smaller cubes. Doing the same with each of these cubes, we get a set consisting of 400 smaller cubes. Continuing this process indefinitely, we get the Menger sponge.

In the theory of functional series, the section devoted to the expansion of a function into a series occupies a central place.

Thus, the problem is posed: for a given function it is required to find such a power series

which converged on some interval and its sum was equal to
, those.

= ..

This task is called the problem of expanding a function into a power series.

A necessary condition for the expansion of a function into a power series is its differentiability an infinite number of times - this follows from the properties of convergent power series. This condition is satisfied, as a rule, for elementary functions in their domain of definition.

So let's assume that the function
has derivatives of any order. Can it be expanded into a power series, if so, how to find this series? The second part of the problem is easier to solve, so let's start with it.

Let's assume that the function
can be represented as the sum of a power series converging in an interval containing a point X 0 :

= .. (*)

where a 0 ,a 1 ,a 2 ,...,a P ,... – uncertain (yet) coefficients.

Let us put in equality (*) the value x = x 0 , then we get

We differentiate the power series (*) term by term

= ..

and putting here x = x 0 , we get

With the next differentiation, we get the series

= ..

assuming x = x 0 , we get
, where
.

After P-fold differentiation we get

Assuming in the last equality x = x 0 , we get
, where

So the coefficients are found

,
,
, …,
,….,

substituting which into a row (*), we get

The resulting series is called near taylor for function
.

Thus, we have established that if the function can be expanded into a power series in powers (x - x 0 ), then this expansion is unique and the resulting series is necessarily a Taylor series.

Note that the Taylor series can be obtained for any function that has derivatives of any order at the point x = x 0 . But this does not yet mean that an equal sign can be put between the function and the resulting series, i.e. that the sum of the series is equal to the original function. Firstly, such an equality can only make sense in the region of convergence, and the Taylor series obtained for the function may diverge, and secondly, if the Taylor series converges, then its sum may not coincide with the original function.

3.2. Sufficient conditions for the expansion of a function into a Taylor series

Let us formulate a statement with the help of which the stated problem will be solved.

If the function
in some neighborhood of the point x 0 has derivatives up to (n+ 1)-th order inclusive, then in this neighborhood we haveformula Taylor

whereR n (X)-residual term of the Taylor formula - has the form (Lagrange form)

where dotξ lies between x and x 0 .

Note that there is a difference between the Taylor series and the Taylor formula: the Taylor formula is a finite sum, i.e. P - fixed number.

Recall that the sum of the series S(x) can be defined as the limit of the functional sequence of partial sums S P (x) at some interval X:

According to this, to expand a function into a Taylor series means to find a series such that for any XX

We write the Taylor formula in the form where

notice, that
defines the error we get, replace the function f(x) polynomial S n (x).

If a
, then
,those. the function expands into a Taylor series. Conversely, if
, then
.

Thus, we have proved criterion for the expansion of a function into a Taylor series.

In order that in some interval the functionf(x) expands in a Taylor series, it is necessary and sufficient that on this interval
, whereR n (x) is the remainder of the Taylor series.

With the help of the formulated criterion, one can obtain sufficientconditions for the expansion of a function into a Taylor series.

If insome neighborhood of the point x 0 the absolute values of all derivatives of a function are limited by the same number M≥ 0, i.e.

, to in this neighborhood, the function expands into a Taylor series.

From the above it follows algorithmfunction expansion f(x) in a Taylor series in the vicinity of the point X 0 :

1. Finding derivative functions f(x):

f(x), f’(x), f”(x), f’”(x), f (n) (x),…

2. We calculate the value of the function and the values of its derivatives at the point X 0

f(x 0 ), f'(x 0 ), f”(x 0 ), f’”(x 0 ), f (n) (x 0 ),…

3. We formally write the Taylor series and find the region of convergence of the resulting power series.

4. We check the fulfillment of sufficient conditions, i.e. establish for which X from the convergence region, remainder term R n (x) tends to zero at
or
.

The expansion of functions in a Taylor series according to this algorithm is called expansion of a function in a Taylor series by definition or direct decomposition.

How to insert mathematical formulas on the site?

You can connect the MathJax library script from a remote server using two code options taken from the main MathJax website or from the documentation page:

Any fractal is built according to a certain rule, which is consistently applied an unlimited number of times. Each such time is called an iteration.

If the function f(x) has on some interval containing a point a, derivatives of all orders, then the Taylor formula can be applied to it:

where r n- the so-called residual term or the remainder of the series, it can be estimated using the Lagrange formula:

, where the number x is enclosed between X and a.

If for some value x r n®0 at n®¥, then in the limit the Taylor formula for this value turns into a convergent formula Taylor series:

So the function f(x) can be expanded into a Taylor series at the considered point X, if:

1) it has derivatives of all orders;

2) the constructed series converges at this point.

At a=0 we get a series called near Maclaurin:

Example 1 f(x)= 2x.

Solution. Let us find the values of the function and its derivatives at X=0

f(x) = 2x, f( 0) = 2 0 =1;

f¢(x) = 2x ln2, f¢( 0) = 2 0 ln2=ln2;

f¢¢(x) = 2x ln 2 2, f¢¢( 0) = 2 0 log 2 2= log 2 2;

f(n)(x) = 2x ln n 2, f(n)( 0) = 2 0 ln n 2=ln n 2.

Substituting the obtained values of the derivatives into the Taylor series formula, we get:

The radius of convergence of this series is equal to infinity, so this expansion is valid for -¥<x<+¥.

Example 2 X+4) for the function f(x)= e x.

Solution. Finding the derivatives of the function e x and their values at the point X=-4.

f(x)= e x, f(-4) = e -4 ;

f¢(x)= e x, f¢(-4) = e -4 ;

f¢¢(x)= e x, f¢¢(-4) = e -4 ;

f(n)(x)= e x, f(n)( -4) = e -4 .

Therefore, the desired Taylor series of the function has the form:

This decomposition is also valid for -¥<x<+¥.

Example 3 . Expand function f(x)=ln x in a series by degrees ( X- 1),

(i.e. in a Taylor series in the vicinity of the point X=1).

Solution. We find the derivatives of this function.

Substituting these values into the formula, we obtain the desired Taylor series:

With the help of d'Alembert's test, one can verify that the series converges when

½ X- 1½<1. Действительно,

The series converges if ½ X- 1½<1, т.е. при 0<x<2. При X=2 we obtain an alternating series that satisfies the conditions of the Leibniz test. At X=0 function is not defined. Thus, the region of convergence of the Taylor series is the half-open interval (0;2].

Let us present the expansions obtained in this way in the Maclaurin series (i.e., in a neighborhood of the point X=0) for some elementary functions:

(2) ,

(3) ,

( the last expansion is called binomial series)

Example 4 . Expand the function into a power series

Solution. In decomposition (1), we replace X on the - X 2 , we get:

Example 5 . Expand the function in a Maclaurin series

Solution. We have

Using formula (4), we can write:

substituting instead of X into the formula -X, we get:

From here we find:

Expanding the brackets, rearranging the terms of the series and making a reduction of similar terms, we get

This series converges in the interval

(-1;1) since it is derived from two series, each of which converges in this interval.

Comment .

Formulas (1)-(5) can also be used to expand the corresponding functions in a Taylor series, i.e. for the expansion of functions in positive integer powers ( Ha). To do this, it is necessary to perform such identical transformations on a given function in order to obtain one of the functions (1) - (5), in which instead of X costs k( Ha) m , where k is a constant number, m is a positive integer. It is often convenient to change the variable t=Ha and expand the resulting function with respect to t in the Maclaurin series.

This method illustrates the theorem on the uniqueness of the expansion of a function in a power series. The essence of this theorem is that in the vicinity of the same point, two different power series cannot be obtained that would converge to the same function, no matter how its expansion is performed.

Example 6 . Expand the function in a Taylor series in a neighborhood of a point X=3.

Solution. This problem can be solved, as before, using the definition of the Taylor series, for which it is necessary to find the derivatives of the functions and their values at X=3. However, it will be easier to use the existing decomposition (5):

The resulting series converges at or -3<x- 3<3, 0<x< 6 и является искомым рядом Тейлора для данной функции.

Example 7 . Write a Taylor series in powers ( X-1) features .

Solution.

The series converges at , or 2< x£5.

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.

Decomposition of non-periodic functions.

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 need to get sine half-cycle Fourier expansion function f(x) in the range from 0 to π, then it is necessary 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 the 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

(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 at half cycle.

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

Portal for the student. Self-training

3.2. Sufficient conditions for the expansion of a function into a Taylor series

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