In more complex structures it may be necessary to repeatedly apply the decomposition theorem. Thus, Figure 2.2 shows the expansion with respect to element 7 (upper row) and then with respect to element 8 (lower row). The resulting four subnets have series-parallel structures and no longer require expansions. It is easy to see that at each step the number of elements in the resulting subnets decreases by one, and the number of subnets requiring further consideration doubles. Therefore, the described process is finite in any case, and the number of resulting series-parallel structures will be 2 m , where t - the number of elements over which the decomposition had to be carried out. The complexity of this method can be estimated as 2 m , which is less than the complexity of exhaustive enumeration, but nevertheless still unacceptable for calculating the reliability of real switching networks.

Figure.2.2 Sequential decomposition of the network

Method of sections or sets of paths

Consider another method for calculating the structural reliability of networks. Suppose, as before, that it is necessary to determine the probability of network connectivity between given pair nodes A,B. The criterion for the correct operation of the network in this case is the presence of at least one way of information transfer between the considered nodes. Suppose we have a list possible ways in the form of a list of elements (nodes and communication directions) included in each path. AT general case paths will be dependent, since any element can be included in several paths. Reliability R s any s-ro path can be calculated using the serial connection formula R s =p 1s p 2s …p ts , where p is - reliability i-th the s-ro element of the path.

The desired reliability of H AB depends on the reliability of each path and the options for their intersections by common elements. Denote the reliability provided by the first r paths, through H r . Adding the next (r+1) -th path with reliability R r+1 will obviously lead to an increase in structural reliability, which will now be determined by the union of two events: at least one of the first r is serviceable paths or serviceable (r+1) - th path. The probability of this combined event occurring, taking into account possible dependence. failures (r+1) - th and other paths

H r+i =H r +R r+i -R r+1 H r/(r+1), (2.10)

where H r/ (r+1) is the probability of serviceability of at least one of the first r paths, provided that the (r+1) -th path is serviceable.

From the definition conditional probability H r/ (r+1) it follows that when calculating it, the probability of correct operation of all elements included in the (r+1) -th path must be taken equal to one. For the convenience of further calculations, we represent the last term of expression (2.10) in the following form:

R r+1 H r/ (r+1) = R r+1 ¤ H r (2.11)

where the symbol (¤) means that when multiplying, the reliability indicators of all elements included in the first r paths and common with the (r+l) -th path are replaced by one. Taking into account (2.11), we can rewrite (2.10):

?H r+1 = R r+1 ¤ Q r (2.12)

where?H r+1 =H r+1 -H r - increment of structural reliability with the introduction of the (r+1) -th path; Q r =1 - H r is the probability that the first r paths will fail simultaneously.

Given that the increase in reliability?H r+1 is numerically equal to the decrease in unreliability?Q r+1, we obtain the following equation in finite differences:

?Q r+1 =R r+1 ¤ Q r (2.13)

It is easy to check that the solution to equation (2.13) is the function

Q r = (1-R 1) ¤ (1-R 2) ¤…¤ (1-R r) ( 2.14)

In the case of independent paths, the operation of symbolic multiplication coincides with ordinary multiplication, and expression (2.14) similarly to (2.4) gives the idle time factor of a system consisting of elements connected in parallel. In the general case, the need to take into account the common elements of the paths forces us to perform multiplication according to (2.14) in algebraic form. In this case, the number of terms in the resulting formula with multiplication by each next binomial is doubled and final result will have 2 r members, which is equivalent to exhaustive enumeration of the totality of all r paths. For example, at r=10, the number of terms in the final formula will exceed 1000, which is already beyond the scope of manual counting. With a further increase in the number of paths, the capabilities of modern computers are quickly exhausted.

However, the properties of the symbolic multiplication operation introduced above make it possible to drastically reduce the complexity of calculations. Let's consider these properties in more detail. According to the operation of symbolic multiplication, the following rule is true for the reliability indicator p i of any element:

p i ¤ p i =p i . (2.15)

Recall that the second factor (2.15) has the meaning of the probability of correct operation of the i-th element under the condition of its serviceability, which, obviously, is equal to one.

To shorten further calculations, we introduce the following notation for the unreliability of the i-th element:

=1-p i (2.16)

Taking into account (2.15) and (2.16), we can write the following simple rules transformations of expressions containing p and p :

p i ¤p i =p i (2.17)

p i p j ¤ =p i p j -p i p s

For an example of the use of these rules in calculating reliability, consider the simplest communication network shown in Fig. Fig.2.3 The letters at the edges of the graph indicate the reliability indicators of the corresponding communication lines.

For simplicity, we will consider nodes to be ideally reliable. Let us assume that for communication between nodes A and B it is possible to use all paths consisting of three or less connected lines in series, i.e. consider the subset of paths (m) = (ab, cdf, cgb, ahf). Let us determine the increment of reliability provided by each subsequent path, according to the formula (2.12) taking into account (2.14):

Зr+1=Rr+1¤ (¤1¤…¤) (2.18),

Figure.2.3 - An example of a calculation network on a limited subset of paths

Figure 2.4 - An example of a network for calculating the reliability of the full set of paths, where Ri=1-R1 is similar to (2.16).

Applying successively the formula (2.18) and the rules of symbolic multiplication (2.17). to the network under consideration, we get

Z 2 =cdf¤ () =cdf*;

Z 3 =cgb¤ (¤) =cgb**;

Z 4 =ahf¤ (¤¤) =ahf**.

When calculating the last increment, we used rule 4, which can be called the rule of absorption of long chains by short ones; in this case, applying it gives b¤cgb=b . If other paths are allowed, such as the cdhb path , then it is not difficult to calculate the reliability increment provided by it?H 5 =cdhb¤ (a¤ f¤ g¤ af) = =cdfb*a*f*g. The resulting network reliability can now be calculated as the sum of the increments provided by each of the considered paths:

H R =?H i (2.19)

So, for the considered example, under the assumption that reliability. all elements of the network is the same, i.e. a=b=c=d=f=h=g=p, we get H 5 =p 2 +p 3 (1-p 2) + +2p 3 (1-p) (1-p 2) +p 4 ( 1-p) 3 . In machine implementation, the calculation can also be based on formula (2.13), taking into account the fact that

Q r =?Q i (2.20)

According to (2.13), we have the following recurrence relation

Q r+i =Q r -R r+1 ¤ Q r . (2.21)

At initial condition Q 0 \u003d l at each subsequent step, from the previously obtained expression for Q r, one should subtract the product of the reliability of the next (r + 1) -th path by the same expression, in which only the reliability indicators of all elements included in the (r + 1) - th path must be set equal to one.

As an example, let's calculate the reliability of the network shown in Figure 2.4 with respect to nodes A and B , between which there are 11 possible ways of information transfer. All calculations are summarized in Table 2.1: a list of elements included in each path, the result of multiplying the reliability of this path by the value of Q r obtained by considering all previous paths, and the result of simplifying the contents of the third column according to the rules (2.17). The final formula for q AB is contained in the last column, read from top to bottom. The table fully shows all the calculations necessary to calculate the structural reliability of the considered network.

Table 2.1 Results of calculating the reliability of the network shown in Fig. 2.4

To reduce the amount of calculations, parentheses should not be unnecessarily opened; if intermediate result allows simplifications (bringing like terms, bracketing the common factor, etc.), they should be performed.

Let us explain several calculation steps. Since Q 0 = 1 (if there are no paths, the network is broken), then for Q 1 from (2.21) Q 1 =1 - ab=ab. We take the next step (6.21) for Q 2 =ab-fghab==ab*fgh and so on.

Let us consider in more detail the step at which the contribution of path 9 is taken into account. The product of the reliability indicators of its constituent elements, recorded in the second column of Table 2.1, is transferred to the third. Next in square brackets the probability of breaking all previous eight paths is written, accumulated in the fourth column (starting from the first row), taking into account the rule (2.15), according to which the reliability indicators of all elements included in path 9 are replaced by ones. The contribution of the fourth, sixth and seventh rows turns out to be equal to zero according to rule 1. Further, the expression in square brackets is simplified according to the rules (2.17) as follows: b =b (fhc-hfc-fhc) =bc (h-fh) =bchf . Similarly, the calculation is made for all other paths.

The use of the method under consideration makes it possible to obtain general formula structural reliability, containing in the considered case only 15 terms instead of the maximum number 2 11 =2048, obtained by directly multiplying the failure probabilities of these paths. In the machine implementation of the method, it is convenient to represent all elements of the network in a positional code as a string of bits and use the built-in Boolean functions to implement the logical elements of transformations (2.17).

So far, we have considered indicators of the structural reliability of the network relative to a dedicated pair of nodes. The totality of such indicators for all or some subset of pairs can quite fully characterize the structural reliability of the network as a whole. Sometimes another, integral, criterion of structural reliability is used. According to this criterion, the network is considered to be serviceable if there is a connection between all its nodes and a requirement is set for the probability of such an event.

To calculate the structural reliability according to this criterion, it is sufficient to introduce a generalization of the concept of a path in the form of a tree connecting all given network nodes. Then the network will be connected if there is at least one connecting tree, and the calculation is reduced to multiplying the failure probabilities of all considered trees, taking into account the presence of common elements. Probability. Q s failure of the s-th tree is defined similarly to the path failure probability

where p is - i-ro reliability indicator of the element included in s-e tree; n s the number of elements in the s-th tree.

Consider, for example, the simplest network in the form of a triangle, sides. which are weighted by reliability indicators a, b, c corresponding branches. For the connectivity of such a network, the existence of at least one of the trees ab, bc, ca is sufficient. . Using the recurrence relation (2.12), we determine the probability that this network is connected H . cb=ab+bca+cab. If a=b=c=p , we get next value connection probability, which is easy to check by enumeration: H . cb \u003d 3r 2 -2r 3.

To calculate the connectivity probability of sufficiently branched networks, instead of the list of connecting trees, as a rule, it is more convenient to use the list of sections (y) that lead to the loss of network connectivity according to the criterion under consideration. It is easy to show that all the rules of symbolic multiplication introduced above are valid for the section, but instead of the reliability indicators of the network elements, the unreliability indicators q=1-p should be used as initial data . Indeed, if all paths or trees can be considered included "in parallel", taking into account their interdependence, then all sections are included in this sense "successively". Let us denote the probability that there is not a single serviceable element in some section s by р s . Then one can write

R s =q 1s q 2s …q ms , (2.22)

where q is - the unreliability index of the i-ro element included in the s-e section.

The probability H cb of network connectivity can then be represented similarly to (2.14) in symbolic form

H cb = (1-p 1 ) ¤ ( 1st 2 ) ¤…¤ ( 1st r) (2.23)

where r - number of considered sections. In other words, in order for the network to be connected, it is necessary that at least one element in each section be operational at the same time, taking into account the mutual dependence of the sections on common elements. Formula (2.23) is in a sense dual to formula (2.14) and is obtained from the latter by replacing paths with cuts and probabilities of good operation with the probability of being in a failure state. Similarly dual with respect to formula (2.21) is the recursive relation

H r+1 =H r - R r+1 ¤ H r (2.24)

For example, let's calculate the connectivity probability of the triangular network considered above with a set of sections ab, bc, ca. According to (2.23) under the initial condition H 0 =1 we have H cd =ab-bca-cab. With the same indicators of unreliability of the network elements a=b=c=q, we get H cb =1-q 2 -2q 2 (1 - q). This result is the same as the one obtained earlier using the tree enumeration method.

The method of sections can, of course, also be used to calculate the probability of network connectivity with respect to a selected pair of nodes, especially in cases where the number of sections in the network under consideration is significant. less than number zeros. However, the greatest effect in terms of reducing the complexity of calculations is given by the simultaneous use of both methods, which will be considered further.

Finding an indefinite integral (a set of antiderivatives or "anti-derivatives") means restoring a function from the known derivative of this function. Restored set of antiderivatives F(x) + With for function f(x) takes into account the integration constant C. By travel speed material point(derivative) the law of motion of this point (primitive) can be restored; according to the acceleration of the movement of a point - its speed and the law of motion. As you can see, integration is a wide field for the activity of Sherlock Holmes from physics. Yes, and in the economy, many concepts are represented through functions and their derivatives, and therefore, for example, it is possible to restore the volume of output produced at the appropriate time by labor productivity at a certain point in time (derivative).

To find the indefinite integral, a fairly small number of basic integration formulas are required. But the process of finding it is much more difficult than the mere application of these formulas. All the complexity does not relate to integration, but to bringing the integrable expression to such a form that makes it possible to find the indefinite integral using the basic formulas mentioned above. This means that to start the practice of integration, you need to activate the results obtained in high school expression transformation skills.

We will learn to find integrals using properties and the table of indefinite integrals from the lesson about the basic concepts of this topic (opens in a new window).

There are several methods for finding the integral, of which variable replacement method and method of integration by parts- a mandatory gentleman's set for everyone who has successfully passed higher mathematics. However, it is more useful and pleasant to start learning integration using the expansion method based on the following two theorems on the properties of the indefinite integral, which we will repeat here for convenience.

Theorem 3. The constant factor in the integrand can be taken out of the sign of the indefinite integral, i.e.

Theorem 4. Indefinite integral of an algebraic sum finite number functions is algebraic sum indefinite integrals of these functions, i.e.

(2)

In addition, the following rule may be useful in integration: if the expression of the integrand contains a constant factor, then the expression of the antiderivative is multiplied by the reciprocal of the constant factor, that is

(3)

Since this lesson is an introduction to solving integration problems, it is important to note two things that are either already initial stage, or a little later may surprise you. Surprise is due to the fact that integration is the inverse operation of differentiation and the indefinite integral can rightly be called "anti-derivative".

The first thing that should not be surprised when integrating. In the table of integrals there are formulas that have no analogues among the formulas of the derivative table . These are the following formulas:

However, one can verify that the derivatives of the expressions on the right-hand sides of these formulas coincide with the corresponding integrands.

The second thing not to be surprised when integrating. Although the derivative of any elementary function is also an elementary function, indefinite integrals of some elementary functions are no longer elementary functions . Examples of such integrals are:

The following skills will be useful to develop an integration technique: reducing fractions, dividing a polynomial in the numerator of a fraction by a monomial in the denominator (to obtain the sum of indefinite integrals), converting roots to a power, multiplying a monomial by a polynomial, raising to a power. These skills are needed to transform the integrand, which should result in the sum of the integrals present in the table of integrals.

Finding indefinite integrals together

Example 1 Find the indefinite integral

.

Decision. We see in the denominator of the integrand a polynomial in which x is squared. This is almost a sure sign that the table integral 21 (with the arc tangent of the result) can be applied. We take out the factor-two from the denominator (there is such a property of the integral - a constant factor can be taken out of the integral sign, it was mentioned above as Theorem 3). The result of all this:

Now the denominator is the sum of squares, which means that we can apply the mentioned table integral. Finally we get the answer:

.

Example 2 Find the indefinite integral

Decision. We again apply Theorem 3 - the property of the integral, on the basis of which the constant factor can be taken out of the integral sign:

We apply formula 7 from the table of integrals (variable in degree) to the integrand:

.

We reduce the resulting fractions and we have the final answer:

Example 3 Find the indefinite integral

Decision. Applying first Theorem 4 and then Theorem 3 on properties, we find this integral as the sum of three integrals:

All three integrals obtained are tabular. We use formula (7) from the table of integrals for n = 1/2, n= 2 and n= 1/5, and then

combines all three arbitrary constants that were introduced when finding three integrals. Therefore, in similar situations, only one arbitrary constant (constant) of integration should be introduced.

Example 4 Find the indefinite integral

Decision. When there is a monomial in the denominator of the integrand, we can divide the numerator by the denominator term by term. The original integral turned into the sum of two integrals:

.

To apply the table integral, we convert the roots to powers and here is the final answer:

We continue to find indefinite integrals together

Example 7 Find the indefinite integral

Decision. If we transform the integrand by squaring the binomial and dividing the numerator by the denominator term by term, then the original integral becomes the sum of three integrals.

This little lesson will allow not only to master typical task, which is quite common in practice, but also to consolidate the materials of the article Expansion of functions into power series. We will need function expansion table in power series , which can be obtained on the page Mathematical formulas and tables. In addition, the reader must understand geometric meaning a definite integral and possess elementary integration skills.

It should also be noted that accuracy up to three decimal places is the most popular. Other accuracy of calculations is also in use, usually 0.01 or 0.0001.

Now the second stage of the solution:
First, we change the integrand to the resulting power series:

Why can this be done at all? This fact explained in class about expansion of functions into power series is an infinite polynomial graph exactly coincides with the graph of the function ! Moreover, in this case, the statement is true for any value of "x", and not just for the integration interval.

In the next step, we simplify each term as much as possible:

It is better to do this right away so that you don’t get confused with unnecessary calculations at the next step.

The calculation technique is standard: first we substitute 0.3 into each term, and then zero. For calculations, we use a calculator:

How many terms of the series must be taken for the final calculations? If the convergent series sign alternating, then absolute error modulo does not exceed the last discarded term of the series. In our case already the third member of the series is less than the required accuracy of 0.001, and therefore if we discard it, then we will certainly make a mistake of no more than 0.000972 (realize why!). Thus, for the final calculation, the first two terms are sufficient: .

Answer: , accurate to 0.001

What did this number come from geometric point vision? is the approximate area of ​​the shaded figure (see figure above).

Example 2

Calculate approximately definite integral, having previously expanded the integrand into a power series, with an accuracy of 0.001

This is an example for independent solution. Complete Solution and the answer at the end of the lesson.

Somehow, undeservedly, I bypassed the arctangent, never putting it in a row. Let's fix the mistake.

Example 3

Calculate the definite integral with an accuracy of 0.01 using the series expansion of the integrand.

Decision: There is a strong suspicion that this integral is taken, however, the solution is not the simplest.

Let us expand the integrand in a Maclaurin series. We use decomposition:

In this case


Here it was lucky that in the end the degrees still remained intact, fractional degrees it would be harder to integrate.

Thus:

It also happens. Members with a cart - it is easier for a student.

Answer: with an accuracy of 0.01.

Again, note that the 0.01 precision is only guaranteed here because the convergent series sign alternating. For a row with positive members, for example, a series such an estimate cannot be made, since the sum of the discarded "tail" can easily exceed 0.00089. What to do in such cases? I'll tell you at the end of the lesson. In the meantime, I will reveal the secret that in all today's examples the rows alternate.

And, of course, should be controlled range of convergence of a series. In the considered example, by the way, it is “cut down”: (due to square root) , but our integration segment lies entirely in this region.

What happens if you try to solve some illegal case like ? The function will also perfectly expand into a series, the members of the series will also be remarkably integrated. But when we start substituting the value upper limit according to the Newton-Leibniz formula, we will see that numbers will grow indefinitely, that is, each next number will be larger than the previous one. The series converges only on the segment . This is not paranoia, in practice it happens from time to time. The reason is a typo in the collection of problems or a training manual, when the authors overlooked that the integration interval “crawls out” beyond the region of convergence of the series.

I will not consider the integral with the arcsine, since it is listed in the Red Book. It is better to additionally consider something "budget":

Example 4

Calculate the definite integral with an accuracy of 0.001 by expanding the integrand into a series and term-by-term integration of this series.

This is a do-it-yourself example. As for zero, it is not a hindrance here - the integrand suffers only repairable gap at the point and therefore improper integral didn’t lie here and nearby, i.e. it is still about definite integral. In the course of the solution, you will see that the resulting series converges beautifully to zero.

In conclusion, let's look at a couple more examples that are somewhat more complicated.

Example 5

Calculate the definite integral with an accuracy of 0.001 by expanding the integrand into a series and term-by-term integration of this series.

Decision: Analyzing the integrand, we come to the conclusion that we need to use the binomial expansion. But first, the function must be represented in the appropriate form:

Unfortunately, none special case binomial expansion is not suitable, and we will have to use the cumbersome general formula:

In this case: ,

It is better to simplify decomposition already at this stage as much as possible. We also note that we obviously do not need the fourth term of the series, since even before integration a fraction appeared in it, which is obviously less than the required accuracy of 0.001.

On the this lesson we will learn how to find integrals of some types of fractions. For successful assimilation of the material, the calculations of articles and should be well understood.

As already noted, in integral calculus there is no convenient formula for integrating a fraction:

And therefore, there is a sad trend: the more “fancy” the fraction, the more difficult it is to find the integral from it. In this regard, we have to resort to various tricks, which we will now discuss.

Numerator decomposition method

Example 1

Find the indefinite integral

Run a check.

On the lesson Indefinite integral. Solution examples we got rid of the product of functions in the integrand, turning it into a sum convenient for integration. It turns out that sometimes a fraction can also be turned into a sum (difference)!

Analyzing the integrand, we notice that both in the numerator and in the denominator we have polynomials of the first degree: x and ( x+3). When the numerator and denominator contain polynomials the same degrees, the following artificial technique helps: in the numerator, we must independently organize the same expression as in the denominator:

.

The reasoning may be as follows: “In the numerator it is necessary to organize ( x+ 3) to bring the integral to the tabular ones, but if I add a triple to the “x”, then, in order for the expression not to change, I must subtract the same triple.

Now we can divide the numerator by the denominator term by term:

As a result, we achieved what we wanted. We use the first two integration rules:

Ready. Check it out yourself if you wish. note that

in the second integral is the "simple" complex function. The features of its integration were discussed in the lesson Variable change method in indefinite integral.

By the way, the considered integral can also be solved by the variable change method, denoting , but the solution will be much longer.

Example 2

Find the indefinite integral

Run a check

This is a do-it-yourself example. It should be noted that here the variable replacement method will no longer work.

Attention important! Examples No. 1, 2 are typical and are common.

In particular, such integrals often arise in the course of solving other integrals, in particular, when integration irrational functions (roots).

The above method also works in the case if the highest power of the numerator is greater than the highest power of the denominator.

Example 3

Find the indefinite integral

Run a check.

Let's start with the numerator. The numerator selection algorithm is something like this:

1) In the numerator we need to organize 2 x-1 but there x 2. What to do? I conclude 2 x-1 in brackets and multiply by x, as: x(2x-1).

2) Now we try to open these brackets, what happens? Get: (2 x 2 -x). Already better, but no deuce at x 2 is not initially in the numerator. What to do? We need to multiply by (1/2), we get:

3) Open the brackets again, we get:

It turned out the right one x 2! But the problem is that an extra term appeared (-1/2) x. What to do? So that the expression does not change, we must add to our construction the same (1/2) x:

. Life has become easier. Is it possible to organize again in the numerator (2 x-1)?

4) You can. We try: . Expand the brackets of the second term:

. Sorry, but we had in the previous step (+1/2) x, not(+ x). What to do? You need to multiply the second term by (+1/2):

.

5) Again, for verification, open the brackets in the second term:

. Now it's okay: received (+1/2) x from the final construction of paragraph 3! But again there is a small “but”, an extra term (-1/4) has appeared, which means we must add (1/4) to our expression:

.

If everything is done correctly, then when opening all the brackets, we should get the initial numerator of the integrand. We check:

It turned out.

Thus:

Ready. In the last term, we applied the method of bringing a function under a differential.

If we find the derivative of the answer and bring the expression to common denominator, then we get exactly the original integrand

Considered decomposition method x 2 in the sum is nothing more than the reverse action to bring the expression to a common denominator.

Algorithm for selecting the numerator in similar examples It's best to do it in draft form. With some skills, it will also work mentally.

In addition to the selection algorithm, you can use the division of a polynomial by a polynomial by a column, but, I'm afraid, the explanations will take up even more space, so some other time.

Example 4

Find the indefinite integral

Run a check.

This is a do-it-yourself example.