Taylor expansion of functions. Expansion of functions into power series

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 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, that is
, then in this interval the function
can be expanded in a Maclaurin series.

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

Solution.

.

,;

,
;

,
;

,

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

,
;

Convergence area
.

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

Solution:

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
.

Solution.

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
.

Solution.

:

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:

Find the interval of convergence 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 .
.

Solution.

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:
.

Solution.

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
.

Solution.

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 this row:

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.

Solution.

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.

.

In this way
.

Example9 . Calculate
with an accuracy of 0.001.

Solution.

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.

Solution.

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.

Solution.

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
, 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?

Decomposition of a function in a series of Taylor, Maclaurin and Laurent on the site for training practical skills. This expansion of a function into a series gives mathematicians an idea of ​​estimating the approximate value of a function at some point in its domain of definition. It is much easier to calculate such a function value, compared to using the Bredis table, which is so out of date in the age of computing. To expand a function into a Taylor series means to calculate the coefficients in front of the linear functions of this series and write it in the correct form. Students confuse these two series, not understanding what is a general case and what is a special case of the second. We remind you once and for all, the Maclaurin series is a special case of the Taylor series, that is, it is the Taylor series, but at the point x = 0. All brief records of the expansion of known functions, such as e^x, Sin(x), Cos(x) and others, these are expansions in a Taylor series, but at the point 0 for the argument. For functions of a complex argument, the Laurent series is the most common problem in the TFKT, since it represents a two-sided infinite series. It is the sum of two rows. We suggest that you look at the decomposition example directly on the website, it is very easy to do this by clicking on the "Example" with any number, and then the "Solution" button. It is to this expansion of a function into a series that the majorizing series is associated, which limits the original function in a certain region along the ordinate axis, if the variable belongs to the abscissa region. Vector analysis comes into comparison with another interesting discipline in mathematics. Since each term needs to be investigated, a lot of time is needed for the process. Any Taylor series can be associated with a Maclaurin series by replacing x0 with zero, but for the Maclaurin series, the reverse representation of the Taylor series is sometimes not obvious. No matter how it is not required to be done in its pure form, it is interesting for general self-development. Each Laurent series corresponds to a two-sided infinite power series in integer powers of z-a, in other words, a series of the same Taylor type, but slightly different in the calculation of the coefficients. We will talk about the region of convergence of the Laurent series a little later, after several theoretical calculations. As in the last century, a phased expansion of a function into a series can hardly be achieved only by reducing the terms to a common denominator, since the functions in the denominators are non-linear. Approximate calculation of the functional value requires the formulation of problems. Think about the fact that when the argument of the Taylor series is a linear variable, then the expansion takes place in several steps, but a completely different picture, when a complex or nonlinear function acts as an argument of the function to be expanded, then the process of representing such a function in a power series is obvious, because, in such a way Thus, it is easy to calculate, albeit approximate, but the value at any point of the domain of definition, with a minimum error that has little effect on further calculations. This also applies to the Maclaurin series. when it is necessary to calculate the function at the zero point. However, the Laurent series itself is here represented by a plane expansion with imaginary units. Also, not without success will be the correct solution of the problem in the course of the overall process. In mathematics, this approach is not known, but it objectively exists. As a result, you can come to the conclusion of the so-called pointwise subsets, and in the expansion of a function in a series, you need to apply methods known for this process, such as applying the theory of derivatives. Once again we are convinced of the correctness of the teacher, who made his assumptions about the results of post-computational calculations. Let's note that the Taylor series, obtained according to all the canons of mathematics, exists and is defined on the entire numerical axis, however, dear users of the website service, do not forget the form of the original function, because it may turn out that it is initially necessary to set the domain of the function, that is, write out and exclude from further considerations those points at which the function is not defined in the domain of real numbers. So to speak, this will show your quickness in solving the problem. The construction of the Maclaurin series with a zero value of the argument will not be an exception to what has been said. At the same time, no one canceled the process of finding the domain of definition of a function, and you must approach this mathematical action with all seriousness. If the Laurent series contains the main part, the parameter "a" will be called an isolated singular point, and the Laurent series will be expanded in the ring - this is the intersection of the areas of convergence of its parts, from which the corresponding theorem will follow. But not everything is as difficult as it may seem at first glance to an inexperienced student. Having studied just the Taylor series, one can easily understand the Laurent series - a generalized case for expanding the space of numbers. Any expansion of a function into a series can only be done at a point in the domain of the function. One should take into account the properties of such functions, for example, periodicity or infinite differentiability. We also suggest that you use the table of ready-made expansions in the Taylor series of elementary functions, since one function can be represented by up to dozens of power series that differ from each other, which can be seen from the use of our online calculator. Maclaurin's online series is easier than ever to determine if you use the unique site service, you just need to enter the correct written function and you will receive the presented answer in a matter of seconds, it will be guaranteed accurate and in a standard written form. You can rewrite the result immediately in a clean copy for delivery to the teacher. It would be correct to first determine the analyticity of the function under consideration in rings, and then unambiguously state that it can be expanded in a Laurent series in all such rings. An important moment is not to lose sight of the members of the Laurent series containing negative degrees. Focus on this as much as possible. Make good use of Laurent's theorem on the expansion of a function into a series in integer powers.

"Find the Maclaurin expansion of f(x)"- this is exactly what the task in higher mathematics sounds like, which some students can do, while others cannot cope with examples. There are several ways to expand a series in powers, here we will give a method for expanding functions in a Maclaurin series. When developing a function in a series, you need to be good at calculating derivatives.

Example 4.7 Expand a function into a series in powers of x

Calculations: We perform the expansion of the function according to the Maclaurin formula. First, we expand the denominator of the function into a series

Finally, we multiply the expansion by the numerator.
The first term is the value of the function at zero f (0) = 1/3.
Find the derivatives of the first and higher order functions f (x) and the value of these derivatives at the point x=0




Further, with the pattern of changing the value of derivatives to 0, we write the formula for the n-th derivative

So, we represent the denominator as an expansion in the Maclaurin series

We multiply by the numerator and get the desired expansion of the function in a series in powers of x

As you can see, there is nothing complicated here.
All key points are based on the ability to calculate derivatives and quickly generalize the value of the derivative of higher orders at zero. The following examples will help you learn how to quickly expand a function into a series.

Example 4.10 Find the Maclaurin expansion of a function

Calculations: As you may have guessed, we will expand the cosine in the numerator in a series. To do this, you can use formulas for infinitesimal values, or you can derive the cosine expansion in terms of derivatives. As a result, we arrive at the next series in powers of x

As you can see, we have a minimum of calculations and a compact representation of the series expansion.

Example 4.16 Expand a function into a series in powers of x:
7/(12-x-x^2)
Calculations: In this kind of examples, it is necessary to expand the fraction through the sum of simple fractions.
How to do this, we will not show now, but with the help of indefinite coefficients we will arrive at the sum of dox fractions.
Next, we write the denominators in exponential form

It remains to expand the terms using the Maclaurin formula. Summing up the terms with the same powers of "x", we compose the formula for the general term of the expansion of the function in a series



The last part of the transition to the series at the beginning is difficult to implement, since it is difficult to combine the formulas for paired and unpaired indices (powers), but with practice you will get better at this.

Example 4.18 Find the Maclaurin expansion of a function

Calculations: Find the derivative of this function:

We expand the function into a series using one of the McLaren formulas:

We summarize the series term by term on the basis that both are absolutely coinciding. By integrating the entire series term by term, we obtain the expansion of the function into a series in powers of x

Between the last two lines of decomposition there is a transition which at the beginning will take you a lot of time. Generalizing the series formula is not easy for everyone, so don't worry about not being able to get a nice and compact formula.

Example 4.28 Find the Maclaurin expansion of the function:

We write the logarithm as follows

Using the Maclaurin formula, we expand the logarithm of the function in a series in powers of x

The final folding is at first glance complicated, but when alternating characters, you will always get something similar. Introductory lesson on the topic of function scheduling in a row is completed. Other no less interesting decomposition schemes will be discussed in detail in the following materials.

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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.

Students of higher mathematics should be aware that the sum of some power series belonging to the interval of convergence of the series given to us is 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. Let us 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 equals any natural number. That is, after making simple calculations, we can come to the conclusion that the series for f (x) \u003d sin x will be of the following form:

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.