Canonical decomposition. What does it mean to factor a number into prime factors? Using divisibility tests for prime factorization

Lecture 14 Random processes Canonical expansion of random processes. Spectral decomposition stationary random process. SluLecture 14
Random processes
Canonical expansion of random processes.
Spectral decomposition of stationary random
process. Random processes with independent
sections. Markov processes and Markov chains.
Normal random processes. Periodically
non-stationary random processes
(Akhmetov S.K.)

Canonical expansion of random processes

Main characteristics of the SP defined by the canonical decomposition

Spectral decomposition of a stationary SP

Spectral decomposition of stationary SP (2)

Spectral decomposition of stationary SP (3)

Random processes with independent cross sections

Markov processes and Markov chains

Normal (Gaussian) random processes

The random variable V is called centered , if its mathematical expectation is equal to 0. An elementary centered random process is the product of a centered random variable V and a non-random function φ(t):X(t)=Vφ(t). An elementary centered random process has the following characteristics:

Expression of the form
, where φ k ( t ), k =1;2;…-non-random functions; , k =1;2;…-uncorrelated centered random variables, called the canonical expansion of the random processX ( t ), while the random variables are called the coefficients of the canonical expansion; and non-random functions φ k ( t ) - coordinate functions of the canonical expansion.

Let's consider the characteristics of a random process

Since by condition
That

Obviously, the same random process has various types canonical expansion depending on the choice of coordinate functions. Moreover, even with the selection of coordinate functions, there is arbitrariness in the distribution of random variables V k. In practice, based on the results of experiments, estimates are obtained for the mathematical expectation and the correlation function:
. After decomposition
into a double Fourier series in coordinate functions φ to (t):

get variance values
random variables V k .

4.2. The concept of a generalized function. Dirac delta function. Integral canonical representation of random processes.

Generalized function is called the limit of a sequence of a one-parameter family of continuous functions.

Dirac delta function
- this is a generalized function resulting from passing to the limit at
in a family of functions

Among the properties -functions we note the following:

2.

3. If f(t)- continuous function, That

Random process X( t ), the correlation function of which has the form called non-stationary “white noise”. If W ( t 1 )= W - const , then X( t )-stationary “white noise”.

As follows from the definition, no two, even however close, sections “ white noise» are not correlated. The expression W(t) is called intensity of "white noise".

Integral canonical representation of the random process X( t ) is called an expression of the form
Where
- random centered function;
- non-random function of continuous arguments

The correlation function of such a random process has the form:

It can be shown that there is a non-random function G(λ) such that

where G(λ 1) is the dispersion density; δ(x) is the Dirac delta function. We get

Therefore, the variance of the random process X(t):

.

4.3. Linear and nonlinear transformations of random processes

The following problem is considered: an “input signal” having the nature of a random process X(t) is supplied to the input of the system (device, converter) S. The system converts it into an “output signal” Y(t):

.

Formally, the transformation of a random process X(t) into Y(t) can be described using the so-called system operator A t:

Y(t)=A t (X(t)).

The index t indicates that this operator performs a time conversion. The following formulations of the problem of transforming a random process are possible.

The laws of distribution are known or general characteristics random process X(t) at the input to the system S, the operator A t of the system S is given, it is necessary to determine the distribution law or general characteristics of the random process Y(t) at the output of the system S.

The laws of distribution (general characteristics) of the random process X(t) and the requirements for the random process Y(t) are known; it is necessary to determine the type of operator A t of the system S that best satisfies the given requirements kY(t).

The laws of distribution (general characteristics) of the random process Y(t) are known and the operator A t of the system S is given; it is required to determine the distribution laws or general characteristics of the random process X(t).

P The following classification of operators A t of the system S is accepted:

System Operators

Linear LNonlinearN

Linear homogeneous L 0 Linear inhomogeneous L n

Let us consider the impact of a linear inhomogeneous system

L n (...)=L 0 (...)+φ(t)

to a random process X(t) having the following canonical expansion:

.

We get:

let us introduce the notation

then the canonical expansion of Y(t) takes the form:

.

Mathematical expectation of the random process Y(t):

correlation function of the random process Y(t):

hence,

On the other side

Variance of the random process Y(t):

In conclusion of this paragraph, we note that the operators of differentiation and integration of random processes are linear homogeneous.

2. The quadratic transformation is considered:

Y(t)=(X(t)) 2 ,

V k -centered random variables having a distribution symmetric about zero; any four of them are jointly independent. Then

Let's introduce non-random functions

and random variables

then the random process Y(t) takes the form

A canonical expansion of the random process Y(t) is obtained. Correlation functionY(t):

In this article you will find all necessary information answering the question how to factor a number into prime factors . First given general idea about the decomposition of a number into prime factors, examples of decompositions are given. Shown further canonical form factoring a number into prime factors. After this, the decomposition algorithm is given arbitrary numbers into prime factors and examples of decomposition of numbers using this algorithm are given. Alternative methods are also considered that allow you to quickly factor small integers into prime factors using divisibility tests and multiplication tables.

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What does it mean to factor a number into prime factors?

First, let's look at what prime factors are.

It is clear that since the word “factors” is present in this phrase, then there is a product of some numbers, and the qualifying word “simple” means that each factor is a prime number. For example, in a product of the form 2·7·7·23 there are four prime factors: 2, 7, 7 and 23.

What does it mean to factor a number into prime factors?

This means that given number must be presented as a product of prime factors, and the value of this product must be equal to the original number. As an example, consider the product of three prime numbers 2, 3 and 5, it is equal to 30, thus the decomposition of the number 30 into prime factors is 2·3·5. Usually the decomposition of a number into prime factors is written as an equality; in our example it will be like this: 30=2·3·5. We emphasize separately that prime factors in the expansion can be repeated. This clearly illustrates next example: 144=2·2·2·2·3·3 . But a representation of the form 45=3·15 is not a decomposition into prime factors, since the number 15 is a composite number.

Arises next question: “What numbers can be factorized into prime factors?”

In search of an answer to it, we present the following reasoning. Prime numbers, by definition, are among those greater than one. Taking into account this fact and , it can be argued that the product of several prime factors is an integer positive number, exceeding one. Therefore, factorization into prime factors occurs only for positive integers that are greater than 1.

But can all integers greater than one be factored into prime factors?

It is clear that it is not possible to factor simple integers into prime factors. This is explained by the fact that prime numbers have only two positive divisors - one and itself, so they cannot be represented as a product of two or more prime numbers. If the integer z could be represented as the product of prime numbers a and b, then the concept of divisibility would allow us to conclude that z is divisible by both a and b, which is impossible due to the simplicity of the number z. However, they believe that any prime number is itself a decomposition.

What about composite numbers? Do they fold out? composite numbers into prime factors, and are all composite numbers subject to such decomposition? The fundamental theorem of arithmetic gives an affirmative answer to a number of these questions. The basic theorem of arithmetic states that any integer a that is greater than 1 can be decomposed into the product of prime factors p 1, p 2, ..., p n, and the decomposition has the form a = p 1 · p 2 ·… · p n, and this the expansion is unique, if you do not take into account the order of the factors

Canonical factorization of a number into prime factors

In the expansion of a number, prime factors can be repeated. Repeating prime factors can be written more compactly using . Let in the decomposition of a number the prime factor p 1 occur s 1 times, the prime factor p 2 – s 2 times, and so on, p n – s n times. Then the prime factorization of the number a can be written as a=p 1 s 1 ·p 2 s 2 ·…·p n s n. This form of recording is the so-called canonical factorization of a number into prime factors.

Let us give an example of the canonical decomposition of a number into prime factors. Let us know the decomposition 609 840=2 2 2 2 3 3 5 7 11 11, its canonical notation has the form 609 840=2 4 3 2 5 7 11 2.

The canonical factorization of a number into prime factors allows you to find all the divisors of the number and the number of divisors of the number.

Algorithm for factoring a number into prime factors

To successfully cope with the task of decomposing a number into prime factors, you need to have a very good knowledge of the information in the article prime and composite numbers.

The essence of the process of decomposing a positive integer number a that exceeds one is clear from the proof of the fundamental theorem of arithmetic. The point is sequential finding the smallest prime divisors p 1 , p 2 , …, p n of the numbers a, a 1 , a 2 , …, a n-1 , which allows us to obtain a series of equalities a = p 1 · a 1 , where a 1 = a: p 1 , a=p 1 ·a 1 =p 1 ·p 2 ·a 2 , where a 2 =a 1:p 2 , …, a=p 1 ·p 2 ·…·p n ·a n , where a n =a n-1 :p n . When we get a n =1, then the equality a=p 1 ·p 2 ·…·p n will give us the desired decomposition of the number a into prime factors. It should also be noted here that p 1 ≤p 2 ≤p 3 ≤…≤p n.

It remains to figure out how to find the smallest prime factors at each step, and we will have an algorithm for decomposing a number into prime factors. A table of prime numbers will help us find prime factors. Let us show how to use it to obtain the smallest prime divisor of the number z.

We sequentially take prime numbers from the table of prime numbers (2, 3, 5, 7, 11, and so on) and divide the given number z by them. The first prime number by which z is evenly divided will be its smallest prime divisor. If the number z is prime, then its smallest prime divisor will be the number z itself. It should also be recalled here that if z is not prime number, then its smallest prime divisor does not exceed the number , where is from z. Thus, if among the prime numbers not exceeding , there was not a single divisor of the number z, then we can conclude that z is a prime number (more about this is written in the theory section under the heading This number is prime or composite).

As an example, we will show how to find the smallest prime divisor of the number 87. Let's take the number 2. Divide 87 by 2, we get 87:2=43 (remaining 1) (if necessary, see article). That is, when dividing 87 by 2, the remainder is 1, so 2 is not a divisor of the number 87. We take the next prime number from the prime numbers table, this is the number 3. Divide 87 by 3, we get 87:3=29. Thus, 87 is divisible by 3, therefore, the number 3 is the smallest prime divisor of the number 87.

Note that in general case To factor the number a into prime factors, we need a table of prime numbers up to a number not less than . We will have to refer to this table at every step, so we need to have it at hand. For example, to factorize the number 95 into prime factors, we will only need a table of prime numbers up to 10 (since 10 is greater than ). And to decompose the number 846,653, you will already need a table of prime numbers up to 1,000 (since 1,000 is greater than ).

We now have enough information to write down algorithm for factoring a number into prime factors. The algorithm for decomposing the number a is as follows:

• Sequentially sorting through the numbers from the table of prime numbers, we find the smallest prime divisor p 1 of the number a, after which we calculate a 1 =a:p 1. If a 1 =1, then the number a is prime, and it itself is its decomposition into prime factors. If a 1 is not equal to 1, then we have a=p 1 ·a 1 and move on to the next step.
• We find the smallest prime divisor p 2 of the number a 1 , to do this we sequentially sort through the numbers from the table of prime numbers, starting with p 1 , and then calculate a 2 =a 1:p 2 . If a 2 =1, then the required decomposition of the number a into prime factors has the form a=p 1 ·p 2. If a 2 is not equal to 1, then we have a=p 1 ·p 2 ·a 2 and move on to the next step.
• Going through the numbers from the table of prime numbers, starting with p 2, we find the smallest prime divisor p 3 of the number a 2, after which we calculate a 3 =a 2:p 3. If a 3 =1, then the required decomposition of the number a into prime factors has the form a=p 1 ·p 2 ·p 3. If a 3 is not equal to 1, then we have a=p 1 ·p 2 ·p 3 ·a 3 and move on to the next step.
• We find the smallest prime divisor p n of the number a n-1 by sorting through the prime numbers, starting with p n-1, as well as a n =a n-1:p n, and a n is equal to 1. This step is the last step of the algorithm; here we obtain the desired decomposition of the number a into prime factors: a=p 1 ·p 2 ·…·p n.

For clarity, all the results obtained at each step of the algorithm for decomposing a number into prime factors are presented in the form of the following table, in which the numbers a, a 1, a 2, ..., a n are written sequentially in a column to the left of the vertical line, and to the right of the line - the corresponding smallest prime divisors p 1, p 2, ..., p n.

All that remains is to consider a few examples of the application of the resulting algorithm for decomposing numbers into prime factors.

Examples of prime factorization

Now we will look in detail examples of factoring numbers into prime factors. When decomposing, we will use the algorithm from the previous paragraph. Let's start with simple cases, and we will gradually complicate them in order to encounter all the possible nuances that arise when decomposing numbers into prime factors.

Example.

Factor the number 78 into its prime factors.

Solution.

We start searching for the first smallest prime divisor p 1 numbers a=78 . To do this, we begin to sequentially sort through prime numbers from the table of prime numbers. We take the number 2 and divide 78 by it, we get 78:2=39. The number 78 is divided by 2 without a remainder, so p 1 =2 is the first found prime divisor of the number 78. In this case, a 1 =a:p 1 =78:2=39. So we come to the equality a=p 1 ·a 1 having the form 78=2·39. Obviously, a 1 =39 is different from 1, so we move on to the second step of the algorithm.

Now we are looking for the smallest prime divisor p 2 of the number a 1 =39. We begin enumerating numbers from the table of prime numbers, starting with p 1 =2. Divide 39 by 2, we get 39:2=19 (remaining 1). Since 39 is not evenly divisible by 2, then 2 is not its divisor. Then we take the next number from the table of prime numbers (number 3) and divide 39 by it, we get 39:3=13. Therefore, p 2 =3 is the smallest prime divisor of the number 39, while a 2 =a 1:p 2 =39:3=13. We have the equality a=p 1 ·p 2 ·a 2 in the form 78=2·3·13. Since a 2 =13 is different from 1, we move on to the next step of the algorithm.

Here we need to find the smallest prime divisor of the number a 2 =13. In search of the smallest prime divisor p 3 of the number 13, we will sequentially sort through the numbers from the table of prime numbers, starting with p 2 =3. The number 13 is not divisible by 3, since 13:3=4 (rest. 1), also 13 is not divisible by 5, 7 and 11, since 13:5=2 (rest. 3), 13:7=1 (rest. 6) and 13:11=1 (rest. 2). The next prime number is 13, and 13 is divisible by it without a remainder, therefore, the smallest prime divisor p 3 of 13 is the number 13 itself, and a 3 =a 2:p 3 =13:13=1. Since a 3 =1, this step of the algorithm is the last, and the desired decomposition of the number 78 into prime factors has the form 78=2·3·13 (a=p 1 ·p 2 ·p 3 ).

Answer:

78=2·3·13.

Example.

Express the number 83,006 as a product of prime factors.

Solution.

At the first step of the algorithm for decomposing a number into prime factors, we find p 1 =2 and a 1 =a:p 1 =83,006:2=41,503, from which 83,006=2·41,503.

In the second step, we find out that 2, 3 and 5 are not prime divisors of the number a 1 =41,503, but the number 7 is, since 41,503:7=5,929. We have p 2 =7, a 2 =a 1:p 2 =41,503:7=5,929. Thus, 83,006=2 7 5 929.

The smallest prime divisor of the number a 2 =5 929 is the number 7, since 5 929:7 = 847. Thus, p 3 =7, a 3 =a 2:p 3 =5 929:7 = 847, from which 83 006 = 2·7·7·847.

Next we find that the smallest prime divisor p 4 of the number a 3 =847 is equal to 7. Then a 4 =a 3:p 4 =847:7=121, so 83 006=2·7·7·7·121.

Now we find the smallest prime divisor of the number a 4 =121, it is the number p 5 =11 (since 121 is divisible by 11 and not divisible by 7). Then a 5 =a 4:p 5 =121:11=11, and 83 006=2·7·7·7·11·11.

Finally, the smallest prime divisor of the number a 5 =11 is the number p 6 =11. Then a 6 =a 5:p 6 =11:11=1. Since a 6 =1, this step of the algorithm for decomposing a number into prime factors is the last, and the desired decomposition has the form 83 006 = 2·7·7·7·11·11.

The result obtained can be written as the canonical decomposition of the number into prime factors 83 006 = 2·7 3 ·11 2.

Answer:

83 006=2 7 7 7 11 11=2 7 3 11 2 991 is a prime number. Indeed, it does not have a single prime divisor not exceeding ( can be roughly estimated as , since it is obvious that 991<40 2 ), то есть, наименьшим делителем числа 991 является оно само. Тогда p 3 =991 и a 3 =a 2:p 3 =991:991=1 . Следовательно, искомое разложение числа 897 924 289 на простые множители имеет вид 897 924 289=937·967·991 .

Answer:

897 924 289 = 937 967 991 .

Using divisibility tests for prime factorization

In simple cases, you can decompose a number into prime factors without using the decomposition algorithm from the first paragraph of this article. If the numbers are not large, then to decompose them into prime factors it is often enough to know the signs of divisibility. Let's give examples for clarification.

For example, we need to factor the number 10 into prime factors. From the multiplication table we know that 2·5=10, and the numbers 2 and 5 are obviously prime, so the prime factorization of the number 10 looks like 10=2·5.

Another example. Using the multiplication table, we will factor the number 48 into prime factors. We know that six is ​​eight - forty-eight, that is, 48 ​​= 6·8. However, neither 6 nor 8 are prime numbers. But we know that twice three is six, and twice four is eight, that is, 6=2·3 and 8=2·4. Then 48=6·8=2·3·2·4. It remains to remember that two times two is four, then we get the desired decomposition into prime factors 48 = 2·3·2·2·2. Let's write this expansion in canonical form: 48=2 4 ·3.

But when factoring the number 3,400 into prime factors, you can use the divisibility criteria. The signs of divisibility by 10, 100 allow us to state that 3,400 is divisible by 100, with 3,400=34·100, and 100 is divisible by 10, with 100=10·10, therefore, 3,400=34·10·10. And based on the test of divisibility by 2, we can say that each of the factors 34, 10 and 10 is divisible by 2, we get 3 400=34 10 10=2 17 2 5 2 5. All factors in the resulting expansion are simple, so this expansion is the desired one. All that remains is to rearrange the factors so that they go in ascending order: 3 400 = 2·2·2·5·5·17. Let us also write down the canonical decomposition of this number into prime factors: 3 400 = 2 3 ·5 2 ·17.

When decomposing a given number into prime factors, you can use in turn both the signs of divisibility and the multiplication table. Let's imagine the number 75 as a product of prime factors. The test of divisibility by 5 allows us to state that 75 is divisible by 5, and we obtain that 75 = 5·15. And from the multiplication table we know that 15=3·5, therefore, 75=5·3·5. This is the required decomposition of the number 75 into prime factors.

References.

• Vilenkin N.Ya. and others. Mathematics. 6th grade: textbook for general education institutions.
• Vinogradov I.M. Fundamentals of number theory.
• Mikhelovich Sh.H. Number theory.
• Kulikov L.Ya. and others. Collection of problems in algebra and number theory: Textbook for students of physics and mathematics. specialties of pedagogical institutes.