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 rn- 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.

Decision. 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.

Decision. 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).

Decision. We find the derivatives of this function.

Substituting these values ​​into the formula, we get 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

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

Example 5 . Expand the function in a Maclaurin series

Decision. 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 neighborhood 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.

Decision. 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 .

Decision.

The series converges at , or 2< x£5.

16.1. Expansion of elementary functions in Taylor series and

Maclaurin

Let us show that if an arbitrary function is defined on the set
, in the vicinity of the point
has many derivatives and is the sum of a power series:

then you can find the coefficients of this series.

Substitute in a power series
. Then
.

Find the first derivative of the function
:

At
:
.

For the second derivative we get:

At
:
.

Continuing this procedure n once we get:
.

Thus, we got a power series of the form:

,

which is called near taylor for function
around the point
.

A special case of the Taylor series is Maclaurin series at
:

The remainder of the Taylor (Maclaurin) series is obtained by discarding the main series n the first terms and is denoted as
. Then the function
can be written as a sum n the first members of the series
and the remainder
:,

.

The rest is usually
expressed in different formulas.

One of them is in the Lagrange form:

, where
.
.

Note that in practice the Maclaurin series is used more often. Thus, in order to write the function
in the form of a sum of a power series, it is necessary:

1) find the coefficients of the Maclaurin (Taylor) series;

2) find the region of convergence of the resulting power series;

3) prove that the given series converges to the function
.

Theorem1 (a necessary and sufficient condition for the convergence of the Maclaurin series). Let the convergence radius of the series
. In order for this series to converge in the interval
to function
, it is necessary and sufficient that the following condition is satisfied:
within the specified interval.

Theorem 2. If derivatives of any order of a function
in some interval
limited in absolute value to the same number M, i.e
, then in this interval the function
can be expanded in a Maclaurin series.

Example1 . Expand in a Taylor series around the point
function.

Decision.

.

,;

,
;

,
;

,

.......................................................................................................................................

,
;

Convergence area
.

Example2 . Expand function in a Taylor series around a point
.

Decision:

We find the value of the function and its derivatives at
.

,
;

,
;

...........……………………………

,
.

Substitute these values ​​in a row. We get:

or
.

Let us find the region of convergence of this series. According to the d'Alembert test, the series converges if

.

Therefore, for any this limit is less than 1, and therefore the area of ​​convergence of the series will be:
.

Let us consider several examples of the expansion into the Maclaurin series of basic elementary functions. Recall that the Maclaurin series:

.

converges on the interval
to function
.

Note that to expand the function into a series, it is necessary:

a) find the coefficients of the Maclaurin series for a given function;

b) calculate the radius of convergence for the resulting series;

c) prove that the resulting series converges to the function
.

Example 3 Consider the function
.

Decision.

Let us calculate the value of the function and its derivatives for
.

Then the numerical coefficients of the series have the form:

for anyone n. We substitute the found coefficients in the Maclaurin series and get:

Find the radius of convergence of the resulting series, namely:

.

Therefore, the series converges on the interval
.

This series converges to the function for any values , because on any interval
function and its absolute value derivatives are limited by the number .

Example4 . Consider the function
.

Decision.

:

It is easy to see that even-order derivatives
, and derivatives of odd order. We substitute the found coefficients in the Maclaurin series and get the expansion:

Let us find the interval of convergence of this series. According to d'Alembert:

for anyone . Therefore, the series converges on the interval
.

This series converges to the function
, because all its derivatives are limited to one.

Example5 .
.

Decision.

Let us find the value of the function and its derivatives at
:

Thus, the coefficients of this series:
and
, hence:

Similarly with the previous series, the area of ​​convergence
. The series converges to the function
, because all its derivatives are limited to one.

Note that the function
odd and series expansion in odd powers, function
– even and expansion in a series in even powers.

Example6 . Binomial series:
.

Decision.

Let us find the value of the function and its derivatives at
:

This shows that:

We substitute these values ​​of the coefficients in the Maclaurin series and obtain the expansion of this function in a power series:

Let's find the radius of convergence of this series:

Therefore, the series converges on the interval
. At the limit points at
and
series may or may not converge depending on the exponent
.

The studied series converges on the interval
to function
, that is, the sum of the series
at
.

Example7 . Let us expand the function in a Maclaurin series
.

Decision.

To expand this function into a series, we use the binomial series for
. We get:

Based on the property of power series (a power series can be integrated in the region of its convergence), we find the integral of the left and right parts of this series:

Find the area of ​​convergence of this series:
,

that is, the convergence region of this series is the interval
. Let us determine the convergence of the series at the ends of the interval. At

. This series is a harmonic series, that is, it diverges. At
we get a number series with a common term
.

The Leibniz series converges. Thus, the region of convergence of this series is the interval
.

16.2. Application of power series of powers in approximate calculations

Power series play an extremely important role in approximate calculations. With their help, tables of trigonometric functions, tables of logarithms, tables of values ​​of other functions that are used in various fields of knowledge, for example, in probability theory and mathematical statistics, were compiled. In addition, the expansion of functions in a power series is useful for their theoretical study. The main issue when using power series in approximate calculations is the question of estimating the error when replacing the sum of a series by the sum of its first n members.

Consider two cases:

the function is expanded into an alternating series;

the function is expanded into a constant-sign series.

Calculation using alternating series

Let the function
expanded into an alternating power series. Then, when calculating this function for a specific value we get a number series to which we can apply the Leibniz test. In accordance with this criterion, if the sum of a series is replaced by the sum of its first n members, then the absolute error does not exceed the first term of the remainder of this series, that is:
.

Example8 . Calculate
with an accuracy of 0.0001.

Decision.

We will use the Maclaurin series for
, substituting the value of the angle in radians:

If we compare the first and second members of the series with a given accuracy, then: .

Third expansion term:

less than the specified calculation accuracy. Therefore, to calculate
it suffices to leave two terms of the series, i.e.

.

Thus
.

Example9 . Calculate
with an accuracy of 0.001.

Decision.

We will use the binomial series formula. For this we write
as:
.

In this expression
,

Let's compare each of the terms of the series with the accuracy that is given. It's clear that
. Therefore, to calculate
it suffices to leave three members of the series.

or
.

Calculation using sign-positive series

Example10 . Calculate number with an accuracy of 0.001.

Decision.

In a row for a function
substitute
. We get:

Let us estimate the error that arises when the sum of the series is replaced by the sum of the first members. Let's write down the obvious inequality:

i.e. 2<<3. Используем формулу остаточного члена ряда в форме Лагранжа:
,
.

According to the condition of the problem, you need to find n such that the following inequality holds:
or
.

It is easy to check that when n= 6:
.

Hence,
.

Example11 . Calculate
with an accuracy of 0.0001.

Decision.

Note that to calculate the logarithms, one could apply the series for the function
, but this series converges very slowly and 9999 terms would have to be taken to achieve the given accuracy! Therefore, to calculate logarithms, as a rule, a series for the function is used
, which converges on the interval
.

Compute
with this row. Let be
, then .

Hence,
,

In order to calculate
with a given accuracy, take the sum of the first four terms:
.

The rest of the row
discard. Let's estimate the error. It's obvious that

or
.

Thus, in the series that was used for the calculation, it was enough to take only the first four terms instead of 9999 in the series for the function
.

Questions for self-diagnosis

1. What is a Taylor series?

2. what kind of series did Maclaurin have?

3. Formulate a theorem on the expansion of a function in a Taylor series.

4. Write the expansion in the Maclaurin series of the main functions.

5. Indicate the areas of convergence of the considered series.

6. How to estimate the error in approximate calculations using power series?

Students of higher mathematics should be aware that the sum of a certain power series belonging to the interval of convergence of the series given to us turns out to be a continuous and unlimited number of times differentiated function. The question arises: is it possible to assert that a given arbitrary function f(x) is the sum of some power series? That is, under what conditions can the function f(x) be represented by a power series? The importance of this question lies in the fact that it is possible to approximately replace the function f(x) by the sum of the first few terms of the power series, that is, by a polynomial. Such a replacement of a function by a rather simple expression - a polynomial - is also convenient when solving some problems, namely: when solving integrals, when calculating, etc.

It is proved that for some function f(x), in which derivatives up to the (n + 1)th order, including the last one, can be calculated, in the neighborhood (α - R; x 0 + R) of some point x = α formula:

This formula is named after the famous scientist Brook Taylor. The series that is obtained from the previous one is called the Maclaurin series:

The rule that makes it possible to expand in a Maclaurin series:

1. Determine the derivatives of the first, second, third ... orders.
2. Calculate what the derivatives at x=0 are.
3. Write the Maclaurin series for this function, and then determine the interval of its convergence.
4. Determine the interval (-R;R), where the remainder of the Maclaurin formula

R n (x) -> 0 for n -> infinity. If one exists, the function f(x) in it must coincide with the sum of the Maclaurin series.

Consider now the Maclaurin series for individual functions.

1. So, the first will be f(x) = e x. Of course, according to its features, such a function has derivatives of very different orders, and f (k) (x) \u003d e x, where k equals everything Let us substitute x \u003d 0. We get f (k) (0) \u003d e 0 \u003d 1, k \u003d 1.2 ... Based on the foregoing, the series e x will look like this:

2. The Maclaurin series for the function f(x) = sin x. Immediately clarify that the function for all unknowns will have derivatives, besides f "(x) \u003d cos x \u003d sin (x + n / 2), f "" (x) \u003d -sin x \u003d sin (x +2*n/2)..., f(k)(x)=sin(x+k*n/2), where k is equal to any natural number.That is, by making simple calculations, we can conclude that the series for f(x) = sin x will look like this:

3. Now let's try to consider the function f(x) = cos x. It has derivatives of arbitrary order for all unknowns, and |f (k) (x)| = |cos(x+k*n/2)|<=1, k=1,2... Снова-таки, произведя определенные расчеты, получим, что ряд для f(х) = cos х будет выглядеть так:

So, we have listed the most important functions that can be expanded in the Maclaurin series, but they are supplemented by Taylor series for some functions. Now we will list them. It is also worth noting that Taylor and Maclaurin series are an important part of the practice of solving series in higher mathematics. So, Taylor series.

1. The first will be a row for f-ii f (x) = ln (1 + x). As in the previous examples, given us f (x) = ln (1 + x), we can add a series using the general form of the Maclaurin series. however, for this function, the Maclaurin series can be obtained much more simply. After integrating a certain geometric series, we get a series for f (x) = ln (1 + x) of such a sample:

2. And the second, which will be final in our article, will be a series for f (x) \u003d arctg x. For x belonging to the interval [-1; 1], the expansion is valid:

That's all. This article examined the most commonly used Taylor and Maclaurin series in higher mathematics, in particular, in economic and technical universities.

If the function f(x) has derivatives of all orders on some interval containing the point a, then the Taylor formula can be applied to it:
,
where rn- the so-called residual term or the remainder of the series, it can be estimated using the Lagrange formula:
, where the number x lies between x and a.

f(x)=

at the point x 0 = Number of row elements 3 4 5 6 7

Use expansion of elementary functions e x , cos(x), sin(x), ln(1+x), (1+x) m

Function entry rules:

If for some value X rn→0 at n→∞, then in the limit the Taylor formula turns for this value into the convergent Taylor series:
,
Thus, 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.

For a = 0 we get a series called near Maclaurin:
,
Expansion of the simplest (elementary) functions in the Maclaurin series:
exponential functions
, R=∞
Trigonometric functions
, R=∞
, R=∞
, (-π/2< x < π/2), R=π/2
The function actgx does not expand in powers of x, because ctg0=∞
Hyperbolic functions


Logarithmic functions
, -1
Binomial series
.

Example #1. Expand the function into a power series f(x)= 2x.
Decision. 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. Write a Taylor series in powers ( X+4) for the function f(x)= e x.
Decision. 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 expansion 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).
Decision. We find the derivatives of this function.
f(x)=lnx , , , ,

f(1)=ln1=0, f"(1)=1, f""(1)=-1, f"""(1)=1*2,..., f(n)=(- 1) n-1 (n-1)!
Substituting these values ​​into the formula, we get the desired Taylor series:

With the help of d'Alembert's test, one can verify that the series converges at ½x-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. For x=0 the function is not defined. Thus, the region of convergence of the Taylor series is the half-open interval (0;2].

Example #4. Expand the function in a power series.
Decision. In decomposition (1) we replace x by -x 2, we get:
, -∞

Example number 5. Expand the function in a Maclaurin series .
Decision. We have
Using formula (4), we can write:

substituting instead of x in the formula -x, we get:

From here we find: ln(1+x)-ln(1-x) = -
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 obtained 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 is based on the theorem on the uniqueness of the expansion of a function in a power series. The essence of this theorem is that in the neighborhood 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 No. 5a. Expand the function in a Maclaurin series, indicate the area of ​​convergence.
Decision. First we find 1-x-6x 2 =(1-3x)(1+2x) , .
to elementary:

The fraction 3/(1-3x) can be viewed as the sum of an infinitely decreasing geometric progression with a denominator of 3x if |3x|< 1. Аналогично, дробь 2/(1+2x) как сумму бесконечно убывающей геометрической прогрессии знаменателем -2x, если |-2x| < 1. В результате получим разложение в степенной ряд

with convergence region |x|< 1/3.

Example number 6. Expand the function in a Taylor series in the vicinity of the point x = 3.
Decision. 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

Example number 7. Write a Taylor series in powers (x -1) of the function ln(x+2) .
Decision.


The series converges at , or -2< x < 5.

Example number 8. Expand the function f(x)=sin(πx/4) in a Taylor series around the point x =2.
Decision. Let's make the replacement t=x-2:

Using expansion (3), in which we substitute π / 4 t for x, we get:

The resulting series converges to the given function at -∞< π / 4 t<+∞, т.е. при (-∞Thus,
, (-∞

Approximate calculations using power series

Power series are widely used in approximate calculations. With their help, with a given accuracy, you can calculate the values ​​of roots, trigonometric functions, logarithms of numbers, definite integrals. Series are also used in the integration of differential equations.
Consider the expansion of the function in a power series:

To calculate the approximate value of a function at a given point X, belonging to the region of convergence of the indicated series, the first n members ( n is a finite number), and the remaining terms are discarded:

To estimate the error of the obtained approximate value, it is necessary to estimate the discarded residual r n (x) . For this, the following methods are used:

• if the resulting series is character-alternating, then the following property is used: for an alternating series that satisfies the Leibniz conditions, the absolute value of the remainder of the series does not exceed the first discarded term.
• if the given series is of constant sign, then the series composed of the discarded terms is compared with an infinitely decreasing geometric progression.
• in the general case, to estimate the remainder of the Taylor series, you can use the Lagrange formula: a x ).

Example #1. Compute ln(3) to within 0.01.
Decision. Let's use the decomposition , where x=1/2 (see example 5 in the previous topic):

Let's check if we can discard the remainder after the first three terms of the expansion, for this we evaluate it using the sum of an infinitely decreasing geometric progression:

So we can discard this remainder and get

Example #2. Calculate to the nearest 0.0001.
Decision. Let's use the binomial series. Since 5 3 is the nearest integer cube to 130, it is advisable to represent the number 130 as 130=5 3 +5.



since the fourth term of the obtained sign-alternating series that satisfies the Leibniz test is already less than the required accuracy:
, so it and the terms following it can be discarded.
Many practically necessary definite or improper integrals cannot be calculated using the Newton-Leibniz formula, because its application is associated with finding an antiderivative, often not having an expression in elementary functions. It also happens that finding an antiderivative is possible, but unnecessarily laborious. However, if the integrand is expanded into a power series, and the limits of integration belong to the interval of convergence of this series, then an approximate calculation of the integral with a predetermined accuracy is possible.

Example #3. Calculate the integral ∫ 0 1 4 sin (x) x to within 10 -5 .
Decision. The corresponding indefinite integral cannot be expressed in elementary functions, i.e. is an "impossible integral". The Newton-Leibniz formula cannot be applied here. Let us calculate the integral approximately.
Dividing term by term the series for sin x on the x, we get:

Integrating this series term by term (this is possible, since the limits of integration belong to the interval of convergence of this series), we obtain:

Since the resulting series satisfies the conditions of Leibniz and it is enough to take the sum of the first two terms in order to obtain the desired value with a given accuracy.
Thus, we find
.

Example #4. Calculate the integral ∫ 0 1 4 e x 2 to within 0.001.
Decision.
. Let's check if we can discard the remainder after the second term of the resulting series.
0.0001<0.001. Следовательно, .

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, on the other hand, you constantly use mathematical formulas on your site, then I recommend that you use MathJax, a special JavaScript library that displays mathematical notation in web browsers using MathML, LaTeX, or ASCIIMathML markup.

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

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

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

The easiest way to connect MathJax is in Blogger or WordPress: in the site control panel, add a widget designed to insert third-party JavaScript code, copy the first or second version of the load code presented above into it, and place the widget closer to the beginning of the template (by the way, this is not at all necessary , 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.