To begin with, we will analyze the theory, then we will solve a couple of examples to consolidate the material on the expansion of a fractionally rational function into a sum of simple fractions. Let's take a closer look at method of uncertain coefficients and partial value method, as well as their combinations.

The simplest fractions are often called elementary fractions.

There are the following types of simple fractions:

where A , M , N , a , p , q are numbers, and the discriminant of the denominator in fractions 3) and 4) is less than zero.

They are called fractions of the first, second, third and fourth types, respectively.

Why break down fractions into simple ones?

Let's give a mathematical analogy. Often you have to simplify the form of an expression so that you can carry out some actions with it. So, the representation of a fractionally rational function as a sum of simple fractions is about the same. It is used to expand functions into power series, Laurent series and, of course, to find integrals.

For example, it requires to take integral of a fractionally rational function. After decomposing the integrand into simple fractions, everything reduces to fairly simple integrals

But about integrals in another section.

Example.

Break down a fraction into its simplest.

Solution.

In general, the ratio of polynomials is decomposed into simple fractions if the degree of the polynomial of the numerator is less than the degree of the polynomial in the denominator. Otherwise, the numerator polynomial is first divided by the denominator polynomial, and only then the correct fractionally rational function is decomposed.

Let's perform division by a column (corner):

Therefore, the original fraction will take the form:

Thus, we will decompose into simple fractions

Algorithm of the method of indeterminate coefficients.

Firstly, factorize the denominator.

In our example, everything is simple - we take x out of brackets.


Secondly, the fraction to be expanded is represented as the sum of simple fractions with uncertain coefficients.

Here it is worth considering the types of expressions that you can have in the denominator.

Enough theory, practice is still clearer.

It's time to return to the example. The fraction is decomposed into the sum of the simplest fractions of the first and third types with indefinite coefficients A , B and C .


Thirdly, we bring the resulting sum of simple fractions with indefinite coefficients to a common denominator and group the terms in the numerator with the same powers x.



That is, we arrive at the equation:


For x nonzero, this equality reduces to the equality of two polynomials


And two polynomials are equal if and only if the coefficients at the same powers are the same.

Fourth, we equate the coefficients at the same powers of x.

In this case, we obtain a system of linear algebraic equations with indefinite coefficients as unknowns:


Fifth, we solve the resulting system of equations in any way (if necessary, see the article) that you like, we find indefinite coefficients.


At sixth, write down the answer.

Please, don't be lazy, check your answer by reducing the resulting expansion to a common denominator.

Method of undetermined coefficients is a universal method for decomposing fractions into simple ones.

It is very convenient to use the partial value method if the denominator is a product of linear factors, that is, it looks like

Let's look at an example to show the advantages of this method.

Example.

Expand a fraction to the simplest.

Solution.

Since the degree of the polynomial in the numerator is less than the degree of the polynomial in the denominator, we do not have to divide. We turn to the decomposition of the denominator into factors.

Let's get the x out of the brackets first.

We find the roots of a square trinomial (for example, according to the Vieta theorem):

Therefore, the square trinomial can be written as

That is, the denominator will take the form

With a given denominator, the original fraction is decomposed into the sum of three simple fractions of the first type with indeterminate coefficients:

We reduce the resulting amount to a common denominator, but in the numerator we do not open the brackets and do not give similar ones for A, B and C (at this stage, it is just the difference from the method of indefinite coefficients):

Thus, we came to equality:

And now, to find indefinite coefficients, we begin to substitute in the resulting equality “private values”, at which the denominator goes to zero, that is, x=0, x=2 and x=3 for our example.

At x=0 we have:

At x=2 we have:

At x=3 we have:

Answer:

As you can see, the difference between the method of uncertain coefficients and the method of partial values ​​is only in the way of finding unknowns. These methods can be combined to simplify calculations.

Consider an example.

Example.

Expand a fractionally rational expression into simple fractions.

Solution.

Since the degree of the numerator polynomial is less than the degree of the denominator polynomial and the denominator has already been factorized, the original expression will be represented as a sum of simple fractions of the following form:

We bring to a common denominator:

Equate the numerators.

Obviously, the zeros of the denominator are the values ​​x=1, x=-1 and x=3. We use the method of partial values.

At x=1 we have:

At x=-1 we have:

At x=3 we have:

It remains to find the unknown and

To do this, we substitute the found values ​​into the equality of numerators:

After opening the brackets and reducing similar terms for the same powers of x, we arrive at the equality of two polynomials:

We equate the corresponding coefficients at the same powers, thereby compiling a system of equations for finding the remaining unknowns and . We get a system of five equations with two unknowns:

From the first equation we immediately find , from the second equation

As a result, we obtain an expansion into simple fractions:

Note.

If we immediately decided to apply the method of indefinite coefficients, then we would have to solve a system of five linear algebraic equations with five unknowns. The use of the partial value method made it easy to find the values ​​of three of the five unknowns, which greatly simplified the further solution.

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 ... solving systems of linear equations! Yes Yes! So those who have problems with systems, repeat matrices, determinants and Cramer's method. And for those comrades who have trouble 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)

And so on. 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 trinomialax 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 the 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 FROM 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 FROM. 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 FROM 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.

There 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 FROM. For this I will express FROM 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, it is also possible to express one of the variables 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:

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

So:

OK it's all over Now. 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 FROM. 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.)

Integration of a fractional-rational function.
Method of undetermined coefficients

We continue to work on integrating fractions. We have already considered integrals of some types of fractions in the lesson, and this lesson in a sense can be considered a continuation. To successfully understand the material, basic integration skills are required, so if you have just started studying integrals, that is, you are a teapot, then you need to start with the article Indefinite integral. Solution examples.

Oddly enough, now we will deal not so much with finding integrals as ... solving systems of linear equations. In this connection strongly I recommend visiting the lesson Namely, you need to be well versed in the substitution methods (the “school” method and the method of term-by-term addition (subtraction) of system equations).

What is a fractional rational function? In simple terms, a fractional-rational function is a fraction in the numerator and denominator of which are polynomials or products of polynomials. At the same time, fractions are more sophisticated than those discussed in the article. Integration of some fractions.

Integration of the correct fractional-rational function

Immediately an example and a typical algorithm for solving the integral of a fractional rational function.

Example 1

Step 1. The first thing we ALWAYS do when solving an integral of a rational-fractional function is ask the following question: is the fraction correct? This step is done orally, and now I will explain how:

First look at the numerator and find out senior degree polynomial:

The highest power of the numerator is two.

Now look at the denominator and find out senior degree denominator. The obvious way is to open the brackets and bring like terms, but you can do it easier, in each parenthesis find the highest degree

and mentally multiply: - thus, the highest degree of the denominator is equal to three. It is quite obvious that if we really open the brackets, then we will not get a degree greater than three.

Conclusion: Highest power of the numerator STRICTLY less than the highest power of the denominator, then the fraction is correct.

If in this example the numerator contained a polynomial 3, 4, 5, etc. degree, then the fraction would be wrong.

Now we will consider only proper fractional-rational functions. The case when the degree of the numerator is greater than or equal to the degree of the denominator, we will analyze at the end of the lesson.

Step 2 Let's factorize the denominator. Let's look at our denominator:

Generally speaking, here is already a product of factors, but, nevertheless, we ask ourselves: is it possible to expand something else? The object of torture, of course, will be the square trinomial. We solve the quadratic equation:

The discriminant is greater than zero, which means that the trinomial is indeed factorized:

General rule: EVERYTHING that in the denominator CAN be factored - factorize

Let's start making a decision:

Step 3 Using the method of indefinite coefficients, we expand the integrand into a sum of simple (elementary) fractions. Now it will be clearer.

Let's look at our integrand function:

And, you know, an intuitive thought somehow slips through that it would be nice to turn our large fraction into several small ones. For example, like this:

The question arises, is it even possible to do this? Let's breathe a sigh of relief, the corresponding theorem of mathematical analysis states - IT IS POSSIBLE. Such a decomposition exists and is unique.

There is only one catch, the coefficients we bye we do not know, hence the name - the method of indefinite coefficients.

You guessed it, the subsequent gestures so, do not cackle! will be aimed at just LEARNING them - to find out what they are equal to.

Be careful, I explain in detail once!

So, let's start dancing from:

On the left side we bring the expression to a common denominator:

Now we safely get rid of the denominators (because they are the same):

On the left side, we open the brackets, while we do not touch the unknown coefficients yet:

At the same time, we repeat the school rule for multiplying polynomials. When I was a teacher, I learned to say this rule with a straight face: In order to multiply a polynomial by a polynomial, you need to multiply each term of one polynomial by each term of the other polynomial.

From the point of view of a clear explanation, it is better to put the coefficients in brackets (although I personally never do this in order to save time):

We compose a system of linear equations.
First, we look for senior degrees:

And we write the corresponding coefficients in the first equation of the system:

Well remember the following nuance. What would happen if the right side did not exist at all? Say, would it just show off without any square? In this case, in the equation of the system, it would be necessary to put zero on the right: . Why zero? And because on the right side you can always attribute this very square with zero: If there are no variables or (and) a free term on the right side, then we put zeros on the right sides of the corresponding equations of the system.

We write the corresponding coefficients in the second equation of the system:

And, finally, mineral water, we select free members.

Eh, ... I was joking. Jokes aside - mathematics is a serious science. In our institute group, no one laughed when the assistant professor said that she would scatter the members along a number line and choose the largest of them. Let's get serious. Although ... whoever lives to see the end of this lesson will still smile quietly.

System ready:

We solve the system:

(1) From the first equation, we express and substitute it into the 2nd and 3rd equations of the system. In fact, it was possible to express (or another letter) from another equation, but in this case it is advantageous to express it from the 1st equation, since there the smallest odds.

(2) We present similar terms in the 2nd and 3rd equations.

(3) We add the 2nd and 3rd equations term by term, while obtaining the equality , from which it follows that

(4) We substitute into the second (or third) equation, from which we find that

(5) We substitute and into the first equation, getting .

If you have any difficulties with the methods of solving the system, work them out in class. How to solve a system of linear equations?

After solving the system, it is always useful to make a check - substitute the found values in each equation of the system, as a result, everything should “converge”.

Almost arrived. The coefficients are found, while:

A clean job should look something like this:

As you can see, the main difficulty of the task was to compose (correctly!) and solve (correctly!) a system of linear equations. And at the final stage, everything is not so difficult: we use the properties of the linearity of the indefinite integral and integrate. I draw your attention to the fact that under each of the three integrals we have a “free” complex function, I spoke about the features of its integration in the lesson Variable change method in indefinite integral.

Check: Differentiate the answer:

The original integrand was obtained, which means that the integral was found correctly.
During the verification, it was necessary to bring the expression to a common denominator, and this is not accidental. The method of indefinite coefficients and bringing the expression to a common denominator are mutually inverse actions.

Example 2

Find the indefinite integral.

Let's go back to the fraction from the first example: . It is easy to see that in the denominator all factors are DIFFERENT. The question arises, what to do if, for example, such a fraction is given: ? Here we have degrees in the denominator, or, in mathematical terms, multiple factors. In addition, there is an indecomposable square trinomial (it is easy to verify that the discriminant of the equation is negative, so the trinomial cannot be factored in any way). What to do? The expansion into a sum of elementary fractions will look like with unknown coefficients at the top or some other way?

Example 3

Submit a function

Step 1. Checking if we have a correct fraction
Highest power of the numerator: 2
Highest denominator: 8
, so the fraction is correct.

Step 2 Can anything be factored in the denominator? Obviously not, everything is already laid out. The square trinomial does not expand into a product for the above reasons. Good. Less work.

Step 3 Let us represent a fractional-rational function as a sum of elementary fractions.
In this case, the decomposition has the following form:

Let's look at our denominator:
When decomposing a fractional-rational function into a sum of elementary fractions, three fundamental points can be distinguished:

1) If the denominator contains a “lonely” factor in the first degree (in our case ), then we put an indefinite coefficient at the top (in our case ). Examples No. 1,2 consisted only of such "lonely" factors.

2) If the denominator contains multiple multiplier, then you need to decompose as follows:
- that is, sequentially sort through all the degrees of "x" from the first to the nth degree. In our example, there are two multiple factors: and , take another look at the decomposition I have given and make sure that they are decomposed exactly according to this rule.

3) If the denominator contains an indecomposable polynomial of the second degree (in our case ), then when expanding in the numerator, you need to write a linear function with indefinite coefficients (in our case, with indefinite coefficients and ).

In fact, there is also a 4th case, but I will keep silent about it, since in practice it is extremely rare.

Example 4

Submit a function as a sum of elementary fractions with unknown coefficients.

This is a do-it-yourself example. Full solution and answer at the end of the lesson.
Strictly follow the algorithm!

If you have figured out the principles by which you need to decompose a fractional-rational function into a sum, then you can crack almost any integral of the type under consideration.

Example 5

Find the indefinite integral.

Step 1. Obviously, the fraction is correct:

Step 2 Can anything be factored in the denominator? Can. Here is the sum of cubes . Factoring the denominator using the abbreviated multiplication formula

Step 3 Using the method of indefinite coefficients, we expand the integrand into a sum of elementary fractions:

Note that the polynomial is indecomposable (check that the discriminant is negative), so at the top we put a linear function with unknown coefficients, and not just a single letter.

We bring the fraction to a common denominator:

Let's create and solve the system:

(1) From the first equation, we express and substitute into the second equation of the system (this is the most rational way).

(2) We present similar terms in the second equation.

(3) We add the second and third equations of the system term by term.

All further calculations, in principle, are oral, since the system is simple.

(1) We write down the sum of fractions in accordance with the found coefficients .

(2) We use the linearity properties of the indefinite integral. What happened in the second integral? You can find this method in the last paragraph of the lesson. Integration of some fractions.

(3) Once again we use the properties of linearity. In the third integral, we begin to select a full square (the penultimate paragraph of the lesson Integration of some fractions).

(4) We take the second integral, in the third we select the full square.

(5) We take the third integral. Ready.

MINISTRY OF SCIENCE AND EDUCATION OF THE REPUBLIC OF BASHKORTO STAN

GAOU SPO Bashkir College of Architecture and Civil Engineering

Khaliullin Askhat Adelzyanovich,

teacher of mathematics Bashkir

College of Architecture and Civil Engineering

UFA

2014

Introduction ___________________________________________________3

Chapter I. Theoretical aspects of using the method of undetermined coefficients ______________________________________________4

Chapter II. Search for solutions to problems with polynomials by the method of indefinite coefficients _______________________________7

2.1. Factoring a polynomial _____________________ 7

2.2. Tasks with parameters__________________________________ 10

2.3. Solving Equations ____________________________________14

2.4. Functional Equations _____________________________19

Conclusion_________________________________________________23

List of references ____________________________24

Application ________________________________________________25

Introduction.

This work is devoted to the theoretical and practical aspects of introducing the method of indefinite coefficients into the school mathematics course. The relevance of this topic is determined by the following circumstances.

No one will argue with the fact that mathematics as a science does not stand in one place, it develops all the time, new tasks of increased complexity appear, which often causes certain difficulties, since these tasks are usually associated with research. In recent years, such problems have been proposed at school, district and republican mathematical Olympiads, they are also available in the USE versions. Therefore, a special method was required that would allow solving at least some of them most quickly, efficiently and affordably. In this work, the content of the method of indefinite coefficients is presented in an accessible way, which is widely used in a wide variety of areas of mathematics, from questions included in the course of a general education school to its most advanced parts. In particular, applications of the method of indefinite coefficients in solving problems with parameters, fractional rational and functional equations are especially interesting and effective; they can easily interest anyone who is interested in mathematics. The main purpose of the proposed work and the selection of problems is to provide ample opportunities for honing and developing the ability to find short and non-standard solutions.

This work consists of two chapters. The first deals with the theoretical aspects of using

method of uncertain coefficients, in the second - practical and methodological aspects of such use.

The appendix to the work contains the conditions of specific tasks for independent solution.

Chapter I . Theoretical aspects of use method of uncertain coefficients

"Man ... was born to be a master,

master, king of nature, but wisdom,

with which he should rule is not given to him

from birth: it is acquired by learning"

N.I. Lobachevsky

There are various ways and methods for solving problems, but one of the most convenient, most effective, original, elegant and at the same time very simple and understandable to everyone is the method of indefinite coefficients. The method of indefinite coefficients is a method used in mathematics to find the coefficients of expressions, the form of which is known in advance.

Before considering the application of the method of indeterminate coefficients to solving various kinds of problems, we present a number of theoretical information.

Let them be given

A n (x) = a 0 x n + a 1 x n-1 + a 2 x n-2 + ··· + a n-1 x + a n

B m (x ) = b 0 x m + b 1 x m -1 + b 2 x m -2 + ··· + b m-1 x + b m ,

polynomials with respect to X with any ratio.

Theorem. Two polynomials depending on one and of the same argument are identically equal if and only ifn = m and their respective coefficients area 0 = b 0 , a 1 = b 1 , a 2 = b 2 ,··· , a n -1 = b m -1 , a n = b m and t. d.

Obviously, equal polynomials take for all values X the same values. Conversely, if the values ​​of two polynomials are equal for all values X, then the polynomials are equal, that is, their coefficients at the same powersX match.

Therefore, the idea of ​​applying the method of indefinite coefficients to solving problems is as follows.

Let us know that as a result of some transformations, an expression of a certain form is obtained and only the coefficients in this expression are unknown. Then these coefficients are denoted by letters and considered as unknown. Then, a system of equations is compiled to determine these unknowns.

For example, in the case of polynomials, these equations are composed of the condition of equality of the coefficients at the same powers X for two equal polynomials.

We will show the above with the following concrete examples, and we will start with the simplest.

So, for example, on the basis of theoretical considerations, the fraction

can be represented as a sum

, where a , b and c - coefficients to be determined. To find them, we equate the second expression with the first:

=

and getting rid of the denominator and collecting on the left the terms with the same powers X, we get:

(a + b + c )X 2 + ( b - c )x - a = 2X 2 – 5 X– 1

Since the last equality must hold for all values X, then the coefficients at the same powersX right and left should be the same. Thus, three equations are obtained for determining the three unknown coefficients:

a+b+c = 2

b - c = - 5

a= 1 , whence a = 1 , b = - 2 , c = 3

Consequently,

=
,

the validity of this equality is easy to verify directly.

Let's also imagine a fraction

as a + b
+ c
+ d
, where a , b , c and d- unknown rational coefficients. Equate the second expression to the first:

a + b
+ c
+ d
=
or, getting rid of the denominator, taking out, where possible, rational factors from under the signs of the roots and bringing like terms on the left side, we get:

(a- 2 b + 3 c ) + (- a+b +3 d )
+ (a+c - 2 d )
+

+ (b-c + d )
= 1 +
-
.

But such an equality is possible only in the case when the rational terms of both parts and the coefficients at the same radicals are equal. Thus, four equations are obtained for finding unknown coefficients a , b , c and d :

a- 2b + 3c = 1

- a+b +3 d = 1

a+c - 2 d = - 1

b - c + d= 0 , whence a = 0 ; b = - ; c = 0 ; d= , that is
= -
+
.

Chapter II. Search for solutions to problems with polynomials method of uncertain coefficients.

“Nothing contributes to the assimilation of the subject

ta how to act with him in different situations"

Academician B.V. Gnedenko

2. 1. Decomposition of a polynomial into factors.

Methods for factoring polynomials:

1) taking the common factor out of brackets; 2) grouping method; 3) application of basic multiplication formulas; 4) introduction of auxiliary terms; 5) preliminary transformation of a given polynomial with the help of certain formulas; 6) expansion by finding the roots of a given polynomial; 7) parameter introduction method; 8) method of uncertain coefficients.

Problem 1. Decompose the polynomial into real factors X 4 + X 2 + 1 .

Solution. There are no roots among the divisors of the free term of this polynomial. We cannot find the roots of a polynomial by other elementary means. Therefore, it is not possible to perform the required expansion by first finding the roots of this polynomial. It remains to look for a solution to the problem either by introducing auxiliary terms or by the method of indefinite coefficients. It's obvious that X 4 + X 2 + 1 = X 4 + X 3 + X 2 - X 3 - X 2 - X + X 2 + X + 1 =

= X 2 (X 2 + X + 1) - X (X 2 + X + 1) + X 2 + X + 1 =

= (X 2 + X + 1)(X 2 - X + 1).

The resulting square trinomials have no roots, and therefore cannot be decomposed into real linear factors.

The described method is technically simple, but difficult due to its artificiality. Indeed, it is very difficult to come up with the required auxiliary terms. Only a guess helped us to find this expansion. But

There are more reliable ways to solve such problems.

One could proceed as follows: suppose that the given polynomial expands into a product

(X 2 + a X + b )(X 2 + c X + d )

two square trinomials with integer coefficients.

Thus, we will have that

X 4 + X 2 + 1 = (X 2 + a X + b )(X 2 + c X + d )

It remains to determine the coefficientsa , b , c and d .

Multiplying the polynomials on the right side of the last equality, we get:X 4 + X 2 + 1 = X 4 +

+ (a + c ) X 3 + (b + a c + d ) X 2 + (ad + bc ) x + bd .

But since we need the right side of this equality to turn into the same polynomial that is on the left side, we require the following conditions to be met:

a + c = 0

b + a c + d = 1

ad + bc = 0

bd = 1 .

The result is a system of four equations with four unknownsa , b , c and d . It is easy to find coefficients from this systema = 1 , b = 1 , c = -1 and d = 1.

Now the problem is solved completely. We got:

X 4 + X 2 + 1 = (X 2 + X + 1)(X 2 - X + 1).

Problem 2. Decompose the polynomial into real factors X 3 – 6 X 2 + 14 X – 15 .

Solution. We represent this polynomial in the form

X 3 – 6 X 2 + 14 X – 15 = (X + a )(X 2 + bx + c) , where a , b and With - not yet determined coefficients. Since two polynomials are identically equal if and only if the coefficients at the same powersX are equal, then, equating the coefficients, respectively, atX 2 , X and free terms, we get a system of three equations with three unknowns:

a+b= - 6

ab+c = 14

ac = - 15 .

The solution of this system will be greatly simplified if we take into account that the number 3 (the divisor of the free term) is the root of this equation, and, therefore,a = - 3 ,

b = - 3 and With = 5 .

Then X 3 – 6 X 2 + 14 X – 15 = (X – 3)(X 2 – 3 x + 5).

The applied method of indefinite coefficients, in comparison with the above method of introducing auxiliary terms, does not contain anything artificial, but on the other hand it requires the application of many theoretical provisions and is accompanied by rather large calculations. For polynomials of higher degree, this method of indeterminate coefficients leads to cumbersome systems of equations.

2.2 Tasks and with parameters.

In recent years, tasks with parameters have been proposed in the USE variants. Their solution often causes certain difficulties. When solving problems with parameters, along with other methods, it is possible to effectively apply the method of indefinite coefficients. It is this method that makes it much easier to solve them and quickly get an answer.

Task 3. Determine at what values ​​of the parameter a equation 2 X 3 – 3 X 2 – 36 X + a – 3 = 0 has exactly two roots.

Solution. 1 way. With the help of a derivative.

We represent this equation in the form of two functions

2x 3 – 3 X 2 – 36 X – 3 = – a .

f (x) = 2x 3 - 3 X 2 – 36 X– 3 and φ( X ) = – a .

Exploring the functionf (x) = 2x 3 - 3 X 2 – 36 X - 3 with the help of a derivative and construct its graph schematically (Fig. 1.).

f( – x )f (x ) , f (– x ) – f (x ). The function is neither even nor odd.

3. Find the critical points of the function, its intervals of increase and decrease, extrema. f / (x ) = 6 x 2 – 6 X – 36. D (f / ) = R , so we find all critical points of the function by solving the equation f / (x ) = 0 .

6(X 2 – X– 6) = 0 ,

X 2 – X– 6 = 0 ,

X 1 = 3 , X 2 = – 2 by the theorem converse to the Vieta theorem.

f / (x ) = 6(X – 3)(X + 2).

+ max - min +

2 3 x

f / (x) > 0 for all X< – 2 and X > 3 and the function is continuous at the pointsx =– 2 and X = 3 , therefore, it increases on each of the intervals (- ; - 2] and [ 3 ; ).

f / (x ) < 0 at - 2 < X< 3 , therefore, it decreases on the interval [- 2; 3 ].

X = - 2 maximum point, because at this point, the sign of the derivative changes from"+" to "-".

f (– 2) = 2 (– 8) – 3 4 – 36 (– 2) – 3 = – 16 – 12 + 72 – 3 == 72 – 31 = 41 ,

x = 3 is the minimum point, since at this point the sign of the derivative changes"-" to "+".

f (3) = 2 27 - 3 9 - 36 3 - 3 = 54 - 27 - 108 - 3 = - 138 + +54 = - 84 .

Graph of the function φ(X ) = – a is a straight line parallel to the x-axis and passing through a point with coordinates (0; – a ). Graphs have two common points at −a= 41 , i.e. a =- 41 and - a= - 84 , i.e. a = 84 .

at

41 φ( X)

2 3 X

3 f ( x ) = 2x 3 – 3 X 2 – 36 X – 3

2 way. Method of uncertain coefficients.

Since, according to the condition of the problem, this equation should have only two roots, the fulfillment of the equality is obvious:

2X 3 – 3 X 2 – 36 X + a – 3 = (x + b ) 2 (2 x + c ) ,

2X 3 – 3 X 2 – 36 X + a – 3 = 2 x 3 + (4 b + c ) x 2 + (2 b 2 + +2 bc ) x + b 2 c ,

Now equating the coefficients at the same powers X, we obtain a system of equations

4 b + c = - 3

2b 2 + 2bc=- 36

b 2 c = a – 3 .

From the first two equations of the system we findb 2 + b – 6 = 0, whence b 1 = - 3 or b 2 = 2 . Respective valuesWith 1 and With 2 it is easy to find from the first equation of the system:With 1 = 9 or With 2 = - 11 . Finally, the desired value of the parameter can be determined from the last equation of the system:

a = b 2 c + 3 , a 1 = - 41 or a 2 = 84.

Answer: this equation has exactly two different

root at a= - 41 and a= 84 .

Task 4. Find the largest value of the parametera , for which the equationX 3 + 5 X 2 + Oh + b = 0

with integer coefficients has three different roots, one of which is - 2 .

Solution. 1 way. Substituting X= - 2 to the left side of the equation, we get

8 + 20 – 2 a + b= 0, which means b = 2 a – 12 .

Since the number - 2 is the root, you can take out the common factor X + 2:

X 3 + 5 X 2 + Oh + b = X 3 + 2 X 2 + 3 X 2 + Oh + (2 a – 12) =

= x 2 (X + 2) + 3 x (X + 2) – 6 x + Oh + (2 a – 12) =

= x 2 (X + 2) + 3 x (X + 2) + (a – 6)(x +2) - 2(a – 6)+ (2 a - 12) =

= (X + 2)(X 2 + 3 x + (a – 6) ) .

By the condition, there are two more roots of the equation. Hence, the discriminant of the second factor is positive.

D =3 2 - 4 (a – 6) = 33 – 4 a > 0 , that is a < 8,25 .

It would seem that the answer would be a = eight . But when substituting the number 8 in the original equation, we get:

X 3 + 5 X 2 + Oh + b = X 3 + 5 X 2 + 8 X + 4 = (X + 2)(X 2 + 3 x + 2 ) =

= (X + 1) (X + 2) 2 ,

that is, the equation has only two distinct roots. But at a = 7 really gets three different roots.

2 way. Method of indefinite coefficients.

If the equation X 3 + 5 X 2 + Oh + b = 0 has a root X = - 2, then you can always pick up numbersc and d so that for allX equality was true

X 3 + 5 X 2 + Oh + b = (X + 2)(X 2 + With x + d ).

For finding numbersc and d open the brackets on the right side, give similar terms and get

X 3 + 5 X 2 + Oh + b = X 3 + (2 + With ) X 2 +(2 with + d ) X + 2 d

Equating the coefficients at the corresponding powers X we have a system

2 + With = 5

2 With + d = a

2 d = b , where c = 3 .

Consequently, X 2 + 3 x + d = 0 , D = 9 – 4 d > 0 or

d < 2.25, so d (- ; 2 ].

The condition of the problem is satisfied by the value d = one . The final desired value of the parametera = 7.

A n e t: when a = 7 this equation has three different roots.

2.3. Solution of equations.

“Remember that when you solve small problems, you

prepare yourself for solving big and difficult

tasks.”

Academician S.L.Sobolev

When solving some equations, it is possible and necessary to show resourcefulness and wit, to apply special techniques. Possession of various methods of transformations and the ability to conduct logical reasoning is of great importance in mathematics. One of these tricks is to add and subtract some well-chosen expression or number. The stated fact itself, of course, is well known to everyone - the main difficulty is to see in a specific configuration those transformations of equations to which it is convenient and expedient to apply it.

On a simple algebraic equation, we illustrate one non-standard method for solving equations.

Problem 5. Solve the equation

=
.

Solution. Multiply both sides of this equation by 5 and rewrite as follows

= 0 ; X 0; -
;

= 0 ,

= 0 ,

= 0 or
= 0

We solve the resulting equations by the method of indefinite coefficients

X 4 - X 3 –7 X – 3 = (X 2 + ah + b )(x 2 + cx + d ) = 0

X 4 - X 3 –7 X – 3 = X 4 + (a + c ) X 3 + (b + a c + d ) X 2 + (ad + bc ) x++ bd

Equating the coefficients at X 3 , X 2 , X and free terms, we get the system

a + c = -1

b + a c + d = 0

ad + bc = -7

bd = -3 , from where we find:a = -2 ; b = - 1 ;

With = 1 ; d = 3 .

So X 4 - X 3 –7X– 3 = (X 2 – 2 X – 1)(X 2 + X + 3) = 0 ,

X 2 – 2 X– 1 = 0 or X 2 + X + 3 = 0

X 1,2 =
no roots.

Similarly, we have

X 4 – 12X – 5 = (X 2 – 2 X – 1)(X 2 + 2X + 5) = 0 ,

where X 2 + 2 X + 5 = 0 , D = - 16 < 0 , нет корней.

Answer: X 1,2 =

Problem 6. Solve the equation

= 10.

Solution. To solve this equation, it is necessary to choose the numbersa and b so that the numerators of both fractions are the same. Therefore, we have a system:

= 0 , X 0; -1 ; -

= - 10

Thus, the task is to pick up the numbersa and b , for which the equality

(a + 6) X 2 + ah- 5 = X 2 + (5 + 2 b ) x + b

Now, according to the theorem on the equality of polynomials, it is necessary that the right side of this equality turns into the same polynomial that is on the left side.

In other words, the relations must hold

a + 6 = 1

a = 5 + 2 b

– 5 = b , from which we find the valuesa = - 5 ;

b = - 5 .

With these valuesa and b equality a + b = - 10 is also valid.

= 0 , X 0; -1 ; -

= 0 ,

= 0 ,

(X 2 – 5X– 5)(X 2 + 3X + 1) = 0 ,

X 2 – 5X– 5 = 0 or X 2 + 3X + 1 = 0 ,

X 1,2 =
, X 3,4 =

Answer: X 1,2 =
, X 3,4 =

Problem 7. Solve the equation

= 4

Solution. This equation is more complicated than the previous ones and therefore we group it in such a way that X 0;-1;3;-8;12

0 ,

= - 4.

From the condition of equality of two polynomials

Oh 2 + (a + 6) X + 12 = X 2 + (b + 11) x – 3 b ,

we obtain and solve the system of equations for unknown coefficientsa and b :

a = 1

a + 6 = b + 11

12 = – 3 b , where a = 1 , b = - 4 .

Polynomials - 3 - 6X + cx 2 + 8 cx and X 2 + 21 + 12 d – dx are identical to each other only when

With = 1

8 With - 6 = - d

3 = 21 + 12 d , With = 1 , d = - 2 .

For valuesa = 1 , b = - 4 , With = 1 , d = - 2

equality
= - 4 is fair.

As a result, this equation takes the following form:

= 0 or
= 0 or
= 0 ,

= - 4 , = - 3 , = 1 , = -
.

From the examples considered, it can be seen how the skillful use of the method of indefinite coefficients,

helps to simplify the solution of a rather complex, unusual equation.

2.4. Functional equations.

“The highest purpose of mathematics ... consists

to find the hidden order in

chaos that surrounds us

N. Wiener

Functional equations are a very general class of equations in which some function is the desired one. A functional equation in the narrow sense of the word is understood as equations in which the desired functions are associated with known functions of one or more variables using the operation of forming a complex function. A functional equation can also be considered as an expression of a property that characterizes a particular class of functions

[ for example, the functional equation f ( x ) = f (- x ) characterizes the class of even functions, the functional equationf (x + 1) = f (x ) is the class of functions with period 1, etc.].

One of the simplest functional equations is the equationf (x + y ) = f (x ) + f (y ). The continuous solutions of this functional equation have the form

f (x ) = Cx . However, in the class of discontinuous functions, this functional equation also has other solutions. The considered functional equation is connected

f (x + y ) = f (x ) · f (y ), f (x y ) = f (x ) + f (y ), f (x y ) = f (x )· f (y ),

continuous solutions, which, respectively, have the form

e cx , FROMlnx , x α (x > 0).

Thus, these functional equations can serve to define exponential, logarithmic and power functions.

The most widely used are equations in whose complex functions the desired ones are external functions. Theoretical and practical applications

it was precisely such equations that prompted eminent mathematicians to study them.

For example, at alignment

f 2 (x) = f (x - y)· f (x + y)

N.I. Lobachevskyused when determining the angle of parallelism in his geometry.

In recent years, problems related to the solution of functional equations are quite often offered at mathematical Olympiads. Their solution does not require knowledge that goes beyond the scope of the mathematics curriculum of general education schools. However, the solution of functional equations often causes certain difficulties.

One of the ways to find solutions to functional equations is the method of indefinite coefficients. It can be used when the appearance of the equation can determine the general form of the desired function. This applies, first of all, to those cases when solutions of equations should be sought among entire or fractional-rational functions.

Let us explain the essence of this technique by solving the following problems.

Task 8. Functionf (x ) is defined for all real x and satisfies for allX R condition

3 f(x) - 2 f(1- x) = x 2 .

Findf (x ).

Solution. Since on the left side of this equation over the independent variable x and the values ​​of the functionf only linear operations are performed, and the right side of the equation is a quadratic function, it is natural to assume that the desired function is also quadratic:

f (X) = ax 2 + bx + c , wherea, b, c – coefficients to be determined, i.e. undetermined coefficients.

Substituting the function into the equation, we arrive at the identity:

3(ax 2 + bx+c) – 2(a(1 – x) 2 + b(1 – x) + c) = x 2 .

ax 2 + (5 b + 4 a) x + (c – 2 a – 2 b) = x 2 .

Two polynomials will be identically equal if they are equal

coefficients at the same powers of the variable:

a = 1

5b + 4a = 0

c– 2 a – 2 b = 0.

From this system we find the coefficients

a = 1 , b = - , c = , alsosatisfiesequality

3 f (x ) - 2 f (1- x ) = x 2 on the set of all real numbers. At the same time, there isx 0 Task 9. Functiony=f(x) for all x is defined, continuous and satisfies the conditionf (f (x)) – f(x) = 1 + 2 x . Find two such functions.

Solution. Two actions are performed on the desired function - the operation of compiling a complex function and

subtraction. Given that the right side of the equation is a linear function, it is natural to assume that the desired function is also linear:f(x) = ax +b , wherea andb are undefined coefficients. Substituting this function intof (f ( (x ) = - X - 1 ;

f 2 (x ) = 2 X+ , which are solutions of the functional equationf (f (x)) – f(x) = 1 + 2 x .

Conclusion.

In conclusion, it should be noted that this work will certainly contribute to the further study of an original and effective method for solving various mathematical problems, which are problems of increased difficulty and require a deep knowledge of the school mathematics course and a high logical culture. Everyone who wants to deepen their knowledge of mathematics on their own will also find in this work, material for reflection and interesting tasks, the solution of which will bring benefit and satisfaction.

In the work, within the framework of the existing school curriculum and in a form accessible for effective perception, the method of indefinite coefficients is presented, which contributes to the deepening of the school mathematics course.

Of course, all the possibilities of the method of indeterminate coefficients cannot be shown in one work. In fact, the method still requires further study and research.

List of used literature.

Glazer G.I. History of mathematics at school.-M.: Education, 1983.

Gomonov S.A. Functional equations in the school course of mathematics // Mathematics at school. - 2000 . -№10 .

Dorofeev G.V., Potapov M.K., Rozov N.Kh.. Manual on mathematics.- M.: Nauka, 1972.

Kurosh A.G. Algebraic Equations of Arbitrary Degrees.-M.: Nauka, 1983.

Likhtarnikov L.M. Elementary introduction to functional equations. - St. Petersburg. : Lan, 1997 .

Manturov O.V., Solntsev Yu.K., Sorokin Yu.I., Fedin N.G. Explanatory dictionary of mathematical terms.-M.: Enlightenment, 1971

Modenov V.P. Mathematics manual. Ch.1.-M.: Moscow State University, 1977.

Modenov V.P. Problems with parameters.-M.: Exam, 2006.

Potapov M.K., Aleksandrov V.V., Pasichenko P.I. Algebra and analysis of elementary functions.- M.: Nauka, 1980.

Khaliullin A.A.. It is possible to solve easier // Mathematics at school. – 2003 . - №8 .

Khaliullin.

4. Expand polynomial 2X 4 – 5X 3 + 9X 2 – 5X+ 3 for multipliers with integer coefficients.

5. At what value a X 3 + 6X 2 + Oh+ 12 on X+ 4 ?

6. At what value of the parametera the equationX 3 +5 X 2 + + Oh + b = 0 with integer coefficients has two different roots, one of which is equal to 1 ?

7. Among the roots of a polynomial X 4 + X 3 – 18X 2 + Oh + b with integer coefficients there are three equal integers. Find the value b .

8. Find the largest integer value of the parameter a, under which the equation X 3 – 8X 2 + ah +b = 0 with integer coefficients has three different roots, one of which is equal to 2.

9. At what values a and b division without remainder X 4 + 3X 3 – 2X 2 + Oh + b on the X 2 – 3X + 2 ?

10. Factorize polynomials:

a)X 4 + 2 X 2 – X + 2 in)X 4 – 4X 3 +9X 2 –8X + 5 e)X 4 + 12X – 5

b)X 4 + 3X 2 + 2X + 3 G)X 4 – 3X –2 e)X 4 – 7X 2 + 1 .

11. Solve the equations:

a)
= 2 = 2 f (1 – X ) = X 2 .

Find f (X) .

13. Function at= f (X) for all X is defined, continuous, and satisfies the condition f ( f (X)) = f (X) + X. Find two such functions.

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 the 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:
;
;
;
;
;
;
;
.

Solution.

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. Function 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.