A rational function is a fraction of the form , whose numerator and denominator are polynomials or products of polynomials.

Example 1 Step 2

.

We multiply indefinite coefficients by polynomials that are not in this individual fraction, but which are in other fractions obtained:

We open the brackets and equate the numerator of the original integrand received to the obtained expression:

In both parts of the equality, we look for terms with the same powers of x and make up a system of equations from them:

.

We cancel all x's and get an equivalent system of equations:

.

Thus, the final expansion of the integrand into the sum of simple fractions:

.

Example 2 Step 2 At step 1, we obtained the following expansion of the original fraction into the sum of simple fractions with indefinite coefficients in the numerators:

.

Now we start looking for uncertain coefficients. To do this, we equate the numerator of the original fraction in the function expression to the numerator of the expression obtained after reducing the sum of fractions to a common denominator:

Now you need to create and solve a system of equations. To do this, we equate the coefficients of the variable to the appropriate degree in the numerator of the original expression of the function and similar coefficients in the expression obtained at the previous step:

We solve the resulting system:

So, from here

.

Example 3 Step 2 At step 1, we obtained the following expansion of the original fraction into the sum of simple fractions with indefinite coefficients in the numerators:

We start looking for uncertain coefficients. To do this, we equate the numerator of the original fraction in the function expression to the numerator of the expression obtained after reducing the sum of fractions to a common denominator:

As in the previous examples, we compose a system of equations:

We reduce x's and get an equivalent system of equations:

Solving the system, we obtain the following values ​​of uncertain coefficients:

We get the final expansion of the integrand into the sum of simple fractions:

.

Example 4 Step 2 At step 1, we obtained the following expansion of the original fraction into the sum of simple fractions with indefinite coefficients in the numerators:

.

How to equate the numerator of the original fraction to the expression in the numerator obtained after decomposing the fraction into the sum of simple fractions and reducing this sum to a common denominator, we already know from the previous examples. Therefore, only for control, we present the resulting system of equations:

Solving the system, we obtain the following values ​​of uncertain coefficients:

We get the final expansion of the integrand into the sum of simple fractions:

Example 5 Step 2 At step 1, we obtained the following expansion of the original fraction into the sum of simple fractions with indefinite coefficients in the numerators:

.

We independently bring this sum to a common denominator, equate the numerator of this expression to the numerator of the original fraction. The result should be the following system of equations:

Solving the system, we obtain the following values ​​of uncertain coefficients:

.

We get the final expansion of the integrand into the sum of simple fractions:

.

Example 6 Step 2 At step 1, we obtained the following expansion of the original fraction into the sum of simple fractions with indefinite coefficients in the numerators:

We perform the same actions with this amount as in the previous examples. The result should be the following system of equations:

Solving the system, we obtain the following values ​​of uncertain coefficients:

.

We get the final expansion of the integrand into the sum of simple fractions:

.

Example 7 Step 2 At step 1, we obtained the following expansion of the original fraction into the sum of simple fractions with indefinite coefficients in the numerators:

.

After known actions with the resulting sum, the following system of equations should be obtained:

Solving the system, we obtain the following values ​​of uncertain coefficients:

We get the final expansion of the integrand into the sum of simple fractions:

.

Example 8 Step 2 At step 1, we obtained the following expansion of the original fraction into the sum of simple fractions with indefinite coefficients in the numerators:

.

Let's make some changes to the actions already brought to automaticity to obtain a system of equations. There is an artificial trick, which in some cases helps to avoid unnecessary calculations. Bringing the sum of fractions to a common denominator, we obtain and equating the numerator of this expression to the numerator of the original fraction, we obtain.

Greetings to all, dear friends!

Well, congratulations! We have safely reached the main material in the integration of rational fractions - method of indeterminate coefficients. Great and mighty.) What is his majesty and power? And it lies in its versatility. It makes sense to know, right? I warn you that there will be several lessons on this topic. For the topic is very long, and the material is extremely important.)

I must say right away that in today's lesson (and subsequent ones too) we will deal not so much with integration as ... systems solution linear equations! Yes Yes! So those who have problems with systems, repeat matrices, determinants and Cramer's method. And for those comrades who have a hard time with matrices, I urge, at worst, to refresh their memory at least "school" methods for solving systems - the substitution method and the term-by-term addition / subtraction method.

To begin our acquaintance, we rewind the film a little back. Let's briefly return to the previous lessons and analyze all those fractions that we have integrated before. Directly, without any method of indeterminate coefficients! Here they are, these fractions. I sorted them into three groups.

Group 1

In the denominator - linear function either on its own or to the extent. In a word, the denominator is the product identical brackets of the form (Ha).

For example:

(x+4) 1 = (x+4)

(x-10) 2 = (x-10)(x-10)

(2x+5) 3 = (2x+5)(2x+5)(2x+5)

Etc. By the way, don't let the parentheses fool you. (4x+5) or (2x+5) 3 with coefficient k inside. It's the same, in essence, brackets of the form (Ha). For this is the most k from such brackets can always be taken out.

Like this:

That's all.) And it doesn't matter what exactly is in the numerator - just dx or some kind of polynomial. We have always expanded the numerator in powers of brackets (x-a), turned a large fraction into a sum of small ones, brought (where necessary) a bracket under the differential and integrated.

Group 2

What do these fractions have in common?

And the common thing is that in all the denominators is square trinomial ax 2 + bx+ c. But not just, namely in a single copy. And it does not matter here whether the discriminant is positive or negative.

Such fractions have always been integrated in one of two ways - either by expanding the numerator in powers of the denominator, or by taking a full square in the denominator and then changing the variable. It all depends on the particular integrand.

Group 3

These were the worst fractions for integrating. The denominator is an indecomposable square trinomial, and even in the degree n. But, again, in a single copy. For, apart from the trinomial, there are no other factors in the denominator. Such fractions are integrated over . Either directly, or reduced to it after selecting the full square in the denominator and then changing the variable.

However, unfortunately, all the rich variety of rational fractions is not limited only to these three considered groups.

But what if the denominator is various parentheses? For example, something like:

(x-1)(x+1)(x+2)

Or at the same time bracket (Ha) and a square trinomial, something like (x-10)(x 2 -2x+17)? And in other similar cases? Here, it is in such cases that it comes to the rescue. method of indeterminate coefficients!

I must say right away: for the time being, we will only work with correct fractions. Those in which the degree of the numerator is strictly less than the degree of the denominator. How to deal with improper fractions is described in detail in fractions. It is necessary to select the whole part (polynomial). By dividing the corner of the numerator by the denominator or by expanding the numerator - as you wish. And even the example is disassembled. And you somehow somehow integrate the polynomial. Not small already go.) But we will also solve examples for improper fractions!

Now let's get to know each other. Unlike most textbooks on higher mathematics, we will not begin our acquaintance with a dry and heavy theory about the fundamental theorem of algebra, Bezout's theorem, about the expansion of a rational fraction into the sum of simplest ones (more on these fractions later) and other tediousness, but we will start with a simple example .

For example, we need to find the following indefinite integral:

First look at the integrand. The denominator is product of three brackets:

(x-1)(x+3)(x+5)

And all brackets various. Therefore, our old technology with the expansion of the numerator in powers of the denominator does not work this time: which bracket should be highlighted in the numerator? (x-1)? (x+3)? It’s not clear ... The selection of the full square in the denominator is also not in the cash register: there is a polynomial third degree (if you multiply all the brackets). What to do?

When looking at our fraction, a completely natural desire arises ... Downright irresistible! From our big fraction, which uncomfortable integrate, somehow make three small ones. At least like this:

Why is this type to be sought? And all because in this form our initial fraction is already comfortable to integrate! Add the denominator of each small fraction and forward.)

Is it even possible to get such a decomposition? The news is good! The corresponding theorem of mathematics says − yes you can! Such a decomposition exists and is unique.

But there is one problem: the coefficients BUT, AT and With we Bye we don't know. And now our main task will be just define them. Find out what our letters are equal to BUT, AT and With. Hence the name, the method uncertain coefficients. Let's start our fabulous journey!

So, we have equality, from which we start dancing:

Let's bring all three fractions to the right to a common denominator and add:

Now you can safely discard the denominators (because they are the same) and simply equate the numerators. Everything is as usual

next step open all brackets(coefficients BUT, AT and With Bye better left outside)

And now (important!) we build our entire structure on the right by seniority: first we collect all members with x 2 in a pile, then - just with x and, finally, we collect free members. In fact, we simply give similar ones and group the terms according to the powers of x.

Like this:

And now we comprehend the result. On the left is our original polynomial. Second degree. The numerator of our integrand. Right too some polynomial of the second degree. Nose unknown coefficients. This equality should be valid for all valid x values. The fractions on the left and on the right were the same (according to our condition)! This means that their numerator and (i.e. our polynomials) are also the same. So the coefficients with the same powers of x these polynomials must have be equal!

We start with the highest degree. From the square. Let's see what kind of coefficients we have at X 2 left and right. On the right we have the sum of the coefficients A+B+C, and on the left - a deuce. So we have the first equation.

We write down:

A+B+C = 2

There is. The first equation is done.)

Then we go along a decreasing trajectory - we look at terms with x in the first degree. On the right at x we ​​have 8A+4B+2C. Good. And what do we have with x on the left? Hm ... On the left, there is no term with X at all! There are only 2x 2 - 3. How to be? Very simple! This means that the coefficient at x on the left we have equals zero! We can write our left side like this:

And what? We have every right.) From here, the second equation looks like this:

8 A+4 B+2 C = 0

Well, practically, that's all. It remains to equate the free terms:

15A-5B-3C = -3

In a word, the equalization of coefficients at the same powers of x occurs according to the following scheme:

All three of our equalities must be satisfied simultaneously. Therefore, we assemble a system from our written equations:

The system is not the most difficult for a diligent student - three equations and three unknowns. Decide as you wish. You can use the Cramer method through matrices with determinants, you can use the Gauss method, you can even use the usual school substitution.

To begin with, I will solve this system in the way that cultural students usually solve such systems. Namely, the Cramer method.

We begin the solution by compiling the system matrix. I remind you that this matrix is ​​just a table made up of coefficients for unknowns.

Here she is:

First of all, we calculate system matrix determinant. Or, briefly, system identifier. It is usually denoted by the Greek letter ∆ ("delta"):

Great, system determinant is not zero (-48≠0) . From the theory of systems of linear equations, this fact means that our system is compatible and has a unique solution.

The next step is to calculate determinants of unknowns ∆A, ∆B, ∆C. I remind you that each of these three determinants is obtained from the main determinant of the system by replacing the columns with coefficients for the corresponding unknowns by a column of free terms.

So we make up the determinants and consider:

I will not explain in detail the technique for calculating third-order determinants here. And don't ask. This is already quite a deviation from the topic will be.) Who is in the subject, he understands what it is about. And, perhaps, you already guessed exactly how I calculated these three determinants.)

That's all and done.)

This is how cultured students usually decide systems. But ... Not all students are friends with determinants. Unfortunately. For some, these simple concepts of higher mathematics forever remain a Chinese letter and a mysterious monster in the fog...

Well, especially for such uncultured students, I propose a more familiar way of solving - method of successive elimination of unknowns. In fact, this is an advanced "school" method of substitution. Only there will be more steps.) But the essence is the same. First of all, I will exclude the variable With. For this I will express With from the first equation and substitute into the second and third:

We simplify, give similar ones and get a new system, already with two unknown:

Now, in this new system, one of the variables can also be expressed in terms of the other. But the most attentive students will probably notice that the coefficients in front of the variable B – opposite. Two and minus two. Therefore, it will be very convenient to add both equations together in order to eliminate the variable AT and leave only the letter BUT.

We add the left and right parts, mentally reduce 2B and -2B and solve the equation only with respect to BUT:

There is. First coefficient found: A = -1/24.

Determine the second coefficient AT. For example, from the top equation:

From here we get:

Fine. The second coefficient is also found: B = -15/8 . There is still a letter left With. To determine it, we use the uppermost equation, where we have it expressed through BUT and AT:

So:

That's it. Unknown odds found! It doesn't matter if it's via Cramer or via substitution. The main thing, right found.)

So, our expansion of a large fraction into a sum of small ones will look like this:

And don't let the resulting fractional coefficients confuse you: in this procedure (the method of indefinite coefficients), this is the most common occurrence. :)

And now it is highly desirable to check whether we have found our coefficients correctly A, B and With. So now we take a draft and remember the eighth grade - we add back all three of our small fractions.

If we get the original large fraction, then everything is fine. No, it means beat me and look for a mistake.

The common denominator will obviously be 24(x-1)(x+3)(x+5).

Go:

Yes!!! Get the original fraction. Which is what needed to be checked. Everything is good. So please don't hit me.)

And now we return to our original integral. It hasn't gotten any easier in that time, yes. But now that our fraction has been decomposed into a sum of small ones, integrating it has become a real pleasure!

See for yourself! We insert our expansion into the original integral.

We get:

We use the properties of linearity and break our large integral into a sum of small ones, we take out all the constants outside the signs of the integral.

We get:

And the resulting three small integrals are already easily taken .

We continue the integration:

That's all.) And don't ask me in this lesson where the logarithms came from in the answer! Who remembers, he is in the subject and will understand everything. And who does not remember - we walk along the links. I don't just put them on.

Final answer:

Here is such a beautiful trinity: three logarithms - a coward, an experienced and a dunce. :) And try, guess such a cunning answer right off the bat! Only the method of indefinite coefficients helps out, yes.) Actually, we are investigating for this purpose. What, how and where.

As a training exercise, I suggest you practice the method and integrate the following fraction:

Practice, find the integral, do not take it for work! You should get an answer like this:

The method of indeterminate coefficients is a powerful thing. It saves even in the most hopeless situation, when you convert the fraction anyway, and so on. And here, some attentive and interested readers may have a number of questions:

- What if the polynomial in the denominator is not factored at all?

- HOW should one look for the expansion of any large rational fraction into a sum of small ones? In any form? Why in this and not that?

- What if there are multiple factors in the expansion of the denominator? Or brackets in powers like (x-1) 2 ? In what form to look for decomposition?

- What if, in addition to simple brackets of the form (x-a), the denominator simultaneously contains an indecomposable square trinomial? Let's say x 2 +4x+5 ? In what form to look for decomposition?

Well, it's time to thoroughly understand where the legs grow from. in the next lesson.)

The method is applicable for minimizing logic algebra functions of any number of variables.

Consider the case of three variables. A Boolean function in a DNF can be represented in the form of all possible conjunctive members that can be included in a DNF:

where kн(0,1) are coefficients. The method consists in selecting the coefficients in such a way that the resulting DNF is minimal.

If we now set all possible values ​​of variables from 000 to 111, then we get 2 n (2 3 =8) equations for determining the coefficients k:

Considering the sets on which the function takes a zero value, determine the coefficients that are equal to 0, and cross them out of the equations, on the right side of which is 1. Of the remaining coefficients in each equation, one coefficient is equated to unity, which determines the conjunction of the smallest rank. The remaining coefficients are equated to 0. So, unit coefficients k determine the corresponding minimum form.

Example. Minimize a given function

if values ​​are known:
;
;
;
;
;
;
;
.

Decision.

After deleting zero coefficients, we get:

=1;

=1;

=1;

=1.

Equate to unity the coefficient , corresponding to the conjunction of the smallest rank and converting the last four equations into 1, and in the first equation it is advisable to equate the coefficient to 1 . The rest of the coefficients are set to 0.

Answer: kind of minimized function .

It should be noted that the method of uncertain coefficients is effective when the number of variables is small and does not exceed 5-6.

Multidimensional cube

Consider a graphical representation of a function in the form of a multidimensional cube. Every vertex n-dimensional cube can be put in correspondence with the unit constituent.

The subset of marked vertices is a mapping onto n-dimensional cube of the Boolean function from n variables in SDNF.

To display the function from n variables presented in any DNF, it is necessary to establish a correspondence between its miniterms and elements n-dimensional cube.

Miniterm (n-1)-th rank
can be considered as the result of gluing two minitherms n-th rank, i.e.

=

On the n-dimensional cube, this corresponds to replacing two vertices that differ only in coordinate values X i connecting these vertices with an edge (the edge is said to cover the vertices incident to it).

Thus, miniterms ( n-1)-th order correspond to the edges of the n-dimensional cube.

Similarly, the correspondence of miniterms ( n-2)-th order faces n-dimensional cube, each of which covers four vertices (and four edges).

Elements n-dimensional cube, characterized by S measurements are called S-cubes.

So vertices are 0-cubes, edges are 1-cubes, faces are 2-cubes, and so on.

Summarizing, we can say that the miniterm ( n-S) rank in DNF for the function n variables is displayed S-cube, and each S-cube covers all those lower-dimensional cubes that are connected only to its vertices.

Example. On fig. given mapping

Here miniterms
and
correspond to 1-cubes ( S=3-2=1), and miniterm X 3 mapped to 2-cubes ( S=3-1=2).

So, any DNF maps to n-dimensional cube set S-cubes that cover all vertices corresponding to the constituents of units (0-cube).

Constituents. For variables X 1 ,X 2 ,…X n expression
is called the constituent of the unit, and
- the constituent of zero ( means either , or ).

This component of unity (zero) turns into unity (zero) only with one set of variable values ​​corresponding to it, which is obtained if all variables are taken equal to one (zero), and their negations - to zero (one).

For example: constituent unit
corresponds to the set (1011), and the zero constituent
- set (1001).

Since SD(K)NF is a disjunction (conjunction) of the constituents of unity (zero), it can be argued that the Boolean function it represents f(x 1 , x 2 ,…, x n) becomes one (zero) only for sets of variable values x 1 , x 2 ,…, x n corresponding to these copies. On other sets, this function turns to 0 (one).

The converse assertion is also true, on which the way of representing as a formula any a boolean function defined by a table.

To do this, it is necessary to write the disjunctions (conjunctions) of the constituents of one (zero) corresponding to the sets of variable values ​​on which the function takes the value equal to one (zero).

For example, the function given by the table

correspond

The resulting expressions can be converted to another form based on the properties of the algebra of logic.

The converse statement is also true: if some set S-cubes covers the set of all vertices corresponding to unit values ​​of the function, then the disjunction corresponding to these S-cubes of miniterms is the expression of the given function in DNF.

It is said that such a set S-cubes (or miniterms corresponding to them) forms a covering of the function. The desire for a minimal form is intuitively understood as a search for such a cover, the number S-cubes of which would be smaller, and their dimension S- more. The cover corresponding to the minimum shape is called the minimum cover.

For example, for the function at=
coverage corresponds to the non-minimum form:

rice a) at=,

a coatings in fig b) at=
, rice c) at=
minimal.

Rice. Feature coverage at=:

a) non-minimal; b), c) minimum.

Function mapping on n-dimensional clearly and simply with n3. A four-dimensional cube can be depicted as shown in Fig., which displays the functions of four variables and its minimum coverage corresponding to the expression at=

Using this method for n>4 requires such complex constructions that it loses all its advantages.

Method of undetermined coefficients

The method is applicable for minimizing logic algebra functions of any number of variables.

Consider the case of three variables. A Boolean function in a DNF can be represented in the form of all possible conjunctive members that can be included in a DNF:

where kО(0,1) are coefficients. The method consists in selecting the coefficients in such a way that the resulting DNF is minimal.

If we now set all possible values ​​of variables from 000 to 111, then we get 2 n (2 3 =8) equations for determining the coefficients k:

Considering the sets on which the function takes a zero value, determine the coefficients that are equal to 0, and delete them from the equations, on the right side of which is 1. Of the remaining coefficients in each equation, one coefficient is equated to one, which determines the conjunction of the smallest rank. The remaining coefficients are equated to 0. So, unit coefficients k determine the corresponding minimum form.

Example. Minimize a given function

if values ​​are known: ; ; ; ; ; ; ; .

Decision.

After deleting zero coefficients, we get:

=1;

=1;

=1.

Let us equate to one the coefficient corresponding to the conjunction of the smallest rank and converting the last four equations into 1, and in the first equation it is advisable to equate the coefficient to 1. The rest of the coefficients are set to 0.

Answer: kind of minimized function .

It should be noted that the method of uncertain coefficients is effective when the number of variables is small and does not exceed 5-6.

Multidimensional cube

Consider a graphical representation of a function in the form of a multidimensional cube. Every vertex n-dimensional cube can be put in correspondence with the unit constituent.

The subset of marked vertices is a mapping onto n-dimensional cube of the Boolean function from n variables in SDNF.

To display the function from n variables presented in any DNF, it is necessary to establish a correspondence between its miniterms and elements n-dimensional cube.

The miniterm of the (n-1)th rank can be considered as the result of gluing two miniterms n-th rank, i.e.

On the n-dimensional cube, this corresponds to replacing two vertices that differ only in coordinate values x i connecting these vertices with an edge (the edge is said to cover the vertices incident to it).

Thus, miniterms ( n-1)-th order correspond to the edges of the n-dimensional cube.

Similarly, the correspondence of miniterms ( n-2)-th order faces n-dimensional cube, each of which covers four vertices (and four edges).

Elements n-dimensional cube, characterized by S measurements are called S-cubes.

So vertices are 0-cubes, edges are 1-cubes, faces are 2-cubes, and so on.

Summarizing, we can say that the miniterm ( n-S) rank in DNF for the function n variables is displayed S-cube, and each S-cube covers all those lower-dimensional cubes that are connected only to its vertices.

Example. On fig. given mapping

Here the miniterms and correspond to 1-cubes ( S=3-2=1), and miniterm x 3 mapped to 2-cubes ( S=3-1=2).

So, any DNF maps to n-dimensional cube set S-cubes that cover all vertices corresponding to the constituents of units (0-cube).

Constituents. For variables x 1,x 2,…x n expression is called the constituent of the unit, and - the constituent of zero (means either , or ).

This component of unity (zero) turns into unity (zero) only with one set of variable values ​​corresponding to it, which is obtained if all variables are taken equal to one (zero), and their negations - to zero (one).

For example: the constituent unit corresponds to the set (1011), and the constituent zero - set (1001).

Since SD(K)NF is a disjunction (conjunction) of the constituents of unity (zero), it can be argued that the Boolean function it represents f(x 1 ,x 2 ,…,x n) becomes one (zero) only for sets of variable values x 1 ,x 2 ,…,x n corresponding to these copies. On other sets, this function turns to 0 (one).

The converse assertion is also true, on which the way of representing as a formula any a boolean function defined by a table.

To do this, it is necessary to write the disjunctions (conjunctions) of the constituents of one (zero) corresponding to the sets of variable values ​​on which the function takes the value equal to one (zero).

For example, the function given by the table

correspond

The resulting expressions can be converted to another form based on the properties of the algebra of logic.

The converse statement is also true: if some set S-cubes covers the set of all vertices corresponding to unit values ​​of the function, then the disjunction corresponding to these S-cubes of miniterms is the expression of the given function in DNF.

It is said that such a set S-cubes (or miniterms corresponding to them) forms a covering of the function. The desire for a minimal form is intuitively understood as a search for such a cover, the number S-cubes of which would be smaller, and their dimension S- more. The cover corresponding to the minimum shape is called the minimum cover.

For example, for the function at= the coverage corresponds to the non-minimum shape.