If the function f(x) has on some interval containing the point A, derivatives of all orders, then the Taylor formula can be applied to it:
Where r n– the so-called remainder term or remainder of the series, it can be estimated using the Lagrange formula:
, where the number x is between X And A.
If for some value x r n®0 at n®¥, then in the limit the Taylor formula turns into a convergent formula for this value Taylor series:
So the function f(x) can be expanded into a Taylor series at the point in question 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 ln 2 2= ln 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 obtain:
The radius of convergence of this series is equal to infinity, therefore this expansion is valid for -¥<x<+¥.
Example 2 X+4) for 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 required Taylor series of the function has the form:
This expansion is also valid for -¥<x<+¥.
Example 3 . Expand a function f(x)=ln x in a series in powers ( X- 1),
(i.e. in the Taylor series in the vicinity of the point X=1).
Solution. Find the derivatives of this function.
Substituting these values into the formula, we obtain the desired Taylor series:
Using d'Alembert's test, you 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 criterion. 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 into the Maclaurin series (i.e. in the vicinity of the point X=0) for some elementary functions:
(2) 
,
(3) 
,
( the last decomposition is called binomial series)
Example 4 . Expand the function into a power series
Solution. In expansion (1) we replace X on - X 2, we get:
Example 5
. Expand the function in a Maclaurin series 
Solution. We have 
Using formula (4), we can write:
substituting instead X into the formula -X, we get:
From here we find:
Opening the brackets, rearranging the terms of the series and bringing 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 into a Taylor series, i.e. for expanding 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 X costs k( Ha) m , where k is a constant number, m is a positive integer. It is often convenient to make a change of 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 a power series expansion of a function. 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 into 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 we need to find the derivatives of the function and their values at X=3. However, it will be easier to use the existing expansion (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 the Taylor series in powers ( X-1) functions 
.
Solution.
The series converges at 
, or 2< x£5.
"Find the Maclaurin series expansion of the function f(x)"- this is exactly what the task in higher mathematics sounds like, which some students can do, while others cannot cope with the examples. There are several ways to expand a series in powers; here we will give a technique for expanding functions into a Maclaurin series. When developing a function in a series, you need to be good at calculating derivatives.
Example 4.7 Expand a function in powers of x
Calculations: We perform the expansion of the function according to the Maclaurin formula. First, let's expand the denominator of the function into a series 
Finally, multiply the expansion by the numerator.
The first term is the value of the function at zero f (0) = 1/3.
Let's find the derivatives of the function of the first and higher orders f (x) and the value of these derivatives at the point x=0 
Next, based on the pattern of changes in the value of derivatives at 0, we write the formula for the nth derivative 
So, we represent the denominator in the form of an expansion in the Maclaurin series 
We multiply by the numerator and obtain 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 higher order derivative at zero. The following examples will help you learn how to quickly arrange a function in a series.
Example 4.10 Find the Maclaurin series expansion of the function
Calculations: As you may have guessed, we will put the cosine in the numerator in a series. To do this, you can use formulas for infinitesimal quantities, or derive the cosine expansion through derivatives. As a result, we arrive at the following 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 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.
We will not show how to do this now, but with the help of indefinite coefficients we will arrive at the sum of fractions.
Next we write the denominators in exponential form 
It remains to expand the terms using the Maclaurin formula. Summing up the terms at the same powers of “x”, we compose a formula for the general term of the expansion of a 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 (degrees), but with practice you will get better at it.
Example 4.18 Find the Maclaurin series expansion of the function
Calculations: Let's find the derivative of this function: 
Let's expand the function into a series using one of McLaren's formulas: 
We sum the series term by term based on the fact that both are absolutely identical. Having integrated the entire series term by term, we obtain the expansion of the function into a series in powers of x 
There is a transition between the last two lines of the expansion which will take a lot of your time at the beginning. Generalizing a series formula isn't easy for everyone, so don't worry about not being able to get a nice, compact formula.
Example 4.28 Find the Maclaurin series expansion of the function:
Let's write the logarithm as follows 
Using Maclaurin’s formula, we expand the logarithm function in a series in powers of x 
The final convolution is complex at first glance, but when alternating signs you will always get something similar. Input lesson on the topic of scheduling functions in a row is completed. Other equally interesting decomposition schemes will be discussed in detail in the following materials.
In the theory of functional series, the central place is occupied by the section devoted to the expansion of a function into a series.
Thus, the task is set: for a given function we need to find such a power series
which converged on a certain 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 decomposability of a function in 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. Is it possible to expand it into a power series? If so, how can we find this series? The second part of the problem is easier to solve, so let’s start with it.
Let us assume that the function 
can be represented as the sum of a power series converging in the interval containing the point X 0 :
=
..
(*)
Where A 0 ,A 1 ,A 2 ,...,A P ,... – unknown (yet) coefficients.
Let us put in equality (*) the value x = x 0 , then we get
.
Let us differentiate the power series (*) term by term
=
..
and believing here x = x 0 , we get
.
With the next differentiation we obtain the series
=
..
believing 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 the series (*), we get
The resulting series is called next to 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 mean that an equal sign can be placed 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 decomposability of a function in a Taylor series
Let us formulate a statement with the help of which the task will be solved.
If the function
in some neighborhood of point x 0 has derivatives up to (n+
1) of order inclusive, then in this neighborhood we haveformula
Taylor
WhereR n (X)-the remainder 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 a 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 XX
Let us write Taylor's formula in the form where
notice, that 
defines the error we get, replace the function f(x)
polynomial S n (x).
If 
, That 
,those. the function is expanded into a Taylor series. Vice versa, if 
, That 
.
Thus we proved criterion for the decomposability of a function in a Taylor series.
In order for the functionf(x) expands into a Taylor series, it is necessary and sufficient that on this interval
, WhereR n (x) is the remainder term of the Taylor series.
Using the formulated criterion, one can obtain sufficientconditions for the decomposability of a function in a Taylor series.
If insome neighborhood of the point x 0 the absolute values of all derivatives of the function are limited to 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 the Taylor series in the vicinity of a point X 0 :
1. Finding derivatives of functions f(x):
f(x), f’(x), f”(x), f’”(x), f (n) (x),…
2. 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. we establish for which X from the convergence region, remainder term R n (x)
tends to zero at 
or
.
The expansion of functions into a Taylor series using this algorithm is called expansion of a function into a Taylor series by definition or direct decomposition.
Expansion of a function into a Taylor, Maclaurin and Laurent series on a site for training practical skills. This series expansion of a function allows mathematicians to estimate the approximate value of the 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 irrelevant in the age of computer technology. To expand a function into a Taylor series means to calculate the coefficients of the linear functions of this series and write it in the correct form. Students confuse these two series, not understanding what is the general case and what is a special case of the second. Let us remind you once and for all that the Maclaurin series is a special case of the Taylor series, that is, this is the Taylor series, but at the point x = 0. All brief entries for the expansion of well-known functions, such as e^x, Sin(x), Cos(x) and others, these are Taylor series expansions, but at point 0 for the argument. For functions of a complex argument, the Laurent series is the most common problem in TFCT, since it represents a two-sided infinite series. It is the sum of two series. We suggest you look at an example of decomposition directly on the website; this is very easy to do by clicking on “Example” with any number, and then the “Solution” button. It is precisely this expansion of a function into a series that is associated with a majorizing series that limits the original function in a certain region along the ordinate axis if the variable belongs to the abscissa region. Vector analysis is compared to another interesting discipline in mathematics. Since each term needs to be examined, the process requires quite a lot of time. Any Taylor series can be associated with a Maclaurin series by replacing x0 with zero, but for a Maclaurin series it is sometimes not obvious to represent the Taylor series in reverse. As if this is not required to be done in its pure form, it is interesting for general self-development. Every 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’ll talk about the region of convergence of the Laurent series a little later, after several theoretical calculations. As in the last century, a step-by-step expansion of a function into a series can hardly be achieved simply by bringing the terms to a common denominator, since the functions in the denominators are nonlinear. An approximate calculation of the functional value is required by the formulation of problems. Think about the fact that when the argument of a Taylor series is a linear variable, then the expansion occurs in several steps, but the picture is completely different when the argument of the function being expanded is a complex or nonlinear function, then the process of representing such a function in a power series is obvious, since, in this way Thus, it is easy to calculate, albeit an approximate value, at any point in the definition region, 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 represented here by an expansion on the plane with imaginary units. Also, the correct solution of the problem during the overall process will not be without success. This approach is not known in mathematics, 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 use methods known for this process, such as the application of the theory of derivatives. Once again we are convinced that the teacher was right when he 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 site service, do not forget the type of the original function, because it may turn out that initially it is necessary to establish the domain of definition of the function, that is, write and exclude from further consideration those points at which the function is not defined in the domain of real numbers. So to speak, this will show your efficiency in solving the problem. The construction of a Maclaurin series with a zero argument value will not be an exception to what has been said. No one has canceled the process of finding the domain of definition of a function, and you must approach this mathematical operation with all seriousness. In the case of a Laurent series containing the main part, the parameter “a” will be called an isolated singular point, and the Laurent series will be expanded in a ring - this is the intersection of the areas of convergence of its parts, hence the corresponding theorem will follow. But not everything is as complicated as it might seem at first glance to an inexperienced student. Having studied the Taylor series, you can easily understand the Laurent series - a generalized case for expanding the space of numbers. Any series expansion of a function can be performed only at a point in the domain of definition of the function. Properties of functions such as periodicity or infinite differentiability should be taken into account. We also suggest that you use the table of ready-made Taylor series expansions of elementary functions, since one function can be represented by up to dozens of different power series, as can be seen from using our online calculator. The online Maclaurin series is as easy as pie to determine, if you use the unique website service, you just need to enter the correct written function and you will receive the presented answer in a matter of seconds, it is guaranteed to be accurate and in a standard written form. You can copy the result directly into a clean copy for submission to the teacher. It would be correct to first determine the analyticity of the function in question in rings, and then unambiguously state that it is expandable in a Laurent series in all such rings. It is important not to lose sight of the terms of the Laurent series containing negative powers. Focus on this as much as possible. Make good use of Laurent's theorem on the expansion of a function in integer powers.