How to decompose a number into a product of prime factors. Taylor expansion Total number of subsets

A special case of the Taylor series at x 0 =0

nwe callnear Maclaurin for functionf(x).

Let us find a decomposition of some elementary functions in the Maclaurin series.

Example 23

.

Solution.

To solve the problem, we will use the algorithm formulated above. Since it is required to expand the function in a Maclaurin series, therefore, we will look for an expansion in the vicinity of the point X 0 = 0.

Let's find the value of the function at the point X 0 =0, derivative functions up to P th order and their values ​​at X 0 = 0:

Let us formally write the Maclaurin series by the formula

Note that we got rad over odd powers, since the coefficients at even powers (when P- an even number) are equal to zero.

Let's find the area of ​​convergence of the obtained series, for this we compose a series of absolute values ​​of the terms of the series:

and apply the sign of D "Alembert to it.

Since the value of the limit does not depend on X and less than unity for any X, then the series converges for all values, which means that the area of ​​convergence of the series X(–,+).

Let us check the fulfillment of sufficient conditions. It's obvious that

For P= 0,1,2,... and for any X,

this means that the function expands into its own Maclaurin series on the entire real axis, i.e.

at X(–,+).

In the considered example, to determine the coefficients of the expansion of a function in a power series in the vicinity of the point X 0 =0 we successively differentiated the function until we could derive a formula for P th derivative, and found the values ​​of the derivatives at a given point. Then it was found out for which X sufficient conditions for the expansion of a function into a series are satisfied. Often these steps lead to cumbersome calculations. These difficulties can sometimes be circumvented by using statement that the expansion of a function in a power series obtained in any way will be its expansion in a Taylor series. Therefore, in order to obtain the expansion of a function in a power series, one can use the already known expansions of elementary functions, the Maclaurin series, applying to them the rules of addition, multiplication of series, and theorems on the integration and differentiation of power series.

For example, the decomposition of the function f(x)= cos x can be obtained by differentiating term-by-term the Maclaurin series expansion of the function f(x) = sin x.

at X(–,+).

Similarly, using the expansion algorithm and theorems on the integration and differentiation of power series, one can obtain Maclaurin series expansions of the following elementary functions:

at X(–,+);

eif t≥.0, or t -1, then the area of ​​​​convergence x (-1;1),

eif–1< T<0 , then the region of convergence x (-1;1].

Such a decomposition is called binomial series. In particular, assuming in the last decomposition T= –1, we get

, X (-1;1).

Replacing in this expansion X to the expression (- X), we get

, at X (–1;1).

Using the power series integration theorem and applying it to the Maclaurin series expansion of the function
, we get

at X (–1;1].

Replacing in the decomposition of functions
variable X to the expression and integrate, we get

At X [–1;1].

Using binomial series - Maclaurin series expansion of the function
, assuming
, replacing X to the expression
and integrating, we get

At X (–1;1).

Example 24.

Using the known expansions, expand the Maclaurin series of the function
.

Solution

It is necessary to find the expansion of the function in the Maclaurin series, i.e. in a power series in powers X. We will use the decomposition

at t  (–1;1].

Assuming t = x 2 , we get

This decomposition is valid when
, where
, then the region of convergence
.

Thus,

Multiplying both sides of the equation by X, we get

at X [–1;1].

Pexample 25

Using known expansions, expand the function
in a Taylor series in the vicinity of the point X 0 =1.

Solution.

It is necessary to obtain the expansion of the function in a Taylor series in the vicinity of the point X 0 = 1, those. by degrees ( X–1).

We will use the decomposition

At t  (-1;1).

In order to obtain the expansion of this function in powers ( X–1) introduce a new variable t= x–1, Then x =t + 1. Let us transform this function to a new variable, setting x =t + 1:

Putting in a known expansion instead of t expression and multiplying by a number, we get

at  (-1;1).

Assuming in the resulting decomposition t = x–1, back to the original variable X and obtain the expansion of this function into a power series in powers ( X-1):

This decomposition is valid under the condition
, where
.

So we got the decomposition

at
.

Example 26

Expand function
in a power series at a point
.

Solution.

We transform this function using the properties of logarithms:

Using the well-known expansion

at t  (–1;1].

find the expansion of the function
, assuming t= 2x, and features
, assuming t= -x:

expansion is valid for 2 X  (–1;1), those. at
.

Likewise,

and the expansion is valid for (– X)  (–1;1), i.e. at X  (–1;1).

Power series can be added term by term and multiplied by a number, which means

moreover, this expansion is valid on the general region of convergence, i.e., at
.

Pexample 27

Expand the function in a Maclaurin series
.

Solution.

Let's transform the function

.

Using the well-known Maclaurin series expansion of the function at=(1+ t) m , assuming
And
, we get

The binomial series used for
has a region of convergence t  (-1;1], therefore, the obtained decomposition is valid for
, where
,
.

So,
at
.

Any natural number can be decomposed into a product of prime factors. If you don't like dealing with large numbers like 5733, learn how to factor them into prime factors (in this case, 3 x 3 x 7 x 7 x 13). A similar task is often encountered in cryptography, which deals with information security problems. If you're not ready to create your own secure email system, learn how to factor numbers into prime factors first.

Steps

Part 1

Finding Prime Factors

1. Start with the original number. Choose a composite number greater than 3. It makes no sense to take a prime number, since it is only divisible by itself and one.

   • Example: we decompose the number 24 into a product of prime numbers.

2. Let us decompose this number into the product of two factors. Find two smaller numbers whose product is equal to the original number. You can use any multiplier, but it's easier to take prime numbers. One good way is to try dividing the original number first by 2, then by 3, then by 5, and see which of those primes it is divisible by.

   • Example: if you don't know the factors for the number 24, try dividing it by small prime numbers. So you will find that the given number is divisible by 2: 24 = 2 x 12. This is a good start.
   • Since 2 is a prime number, it is good to use it when decomposing even numbers.

3. Start building a multiplier tree. This simple procedure will help you factorize a number into prime factors. To get started, draw two "branches" down from the original number. At the end of each branch, write the found multipliers.

   • Example:

4. Factor the following row of numbers. Take a look at the two new numbers (second row of the multiplier tree). Are they both prime numbers? If one of them is not prime, also factor it into two factors. Draw two more branches and write two new multipliers in the third line of the tree.

   • Example: 12 is not a prime number, so it must be factored. We use the decomposition 12 = 2 x 6 and write it in the third line of the tree:
   • 2x6

5. Keep moving down the tree. If one of the new factors turns out to be a prime number, draw one "branch" from it and write the same number at its end. Prime numbers are not decomposed into smaller factors, so just transfer them to the level below.

   • Example: 2 is a prime number. Just move the 2 from the second to the third line:
   • 2 2 6

6. Keep factoring numbers until you are left with only prime numbers. Check each new line of the tree. If at least one of the new factors is not a prime number, factor it and write a new line. In the end, you will be left with only prime numbers.

   • Example: 6 is not a prime number, so it should also be factored. At the same time, 2 is a prime number, and we carry two 2s to the next level:
   • 2 2 6
   • / / /\
   • 2 2 2 3

7. Write the last line as a product of prime factors. In the end, you will be left with only prime numbers. When this happens, the prime factorization is complete. The last line is a set of prime numbers whose product gives the original number.

   • Check your answer: multiply the numbers in the last line. The result should be the original number.
   • Example: The last row of the factor tree contains the numbers 2 and 3. Both of these numbers are prime, so the expansion is complete. Thus, the decomposition of the number 24 into prime factors has the following form: 24 = 2 x 2 x 2 x 3.
   • The order of the multipliers does not matter. The expansion can also be written as 2 x 3 x 2 x 2.

8. If you wish, simplify your answer using power notation. If you are familiar with raising numbers to a power, you can write your answer in a simpler way. Remember that the base is written below, and the superscript number shows how many times this base should be multiplied by itself.

   • Example: how many times does the number 2 appear in the found expansion of 2 x 2 x 2 x 3? Three times, so the expression 2 x 2 x 2 can be written as 2 3 . In simplified notation, we get 23x3.

   Part 2

   Using Prime Factorization

   1. Find the greatest common divisor of two numbers. The greatest common divisor (GCD) of two numbers is the maximum number by which both numbers are divisible without a remainder. The following example shows how to use prime factorization to find the greatest common divisor of 30 and 36.

      • Let's decompose both numbers into prime factors. For the number 30, the expansion is 2 x 3 x 5. The number 36 is decomposed into prime factors as follows: 2 x 2 x 3 x 3.
      • Find a number that occurs in both expansions. We cross out this number in both lists and write it on a new line. For example, 2 occurs in two expansions, so we write 2 in a new line. After that, we are left with 30 = 2 x 3 x 5 and 36 = 2 x 2 x 3 x 3.
      • Repeat this action until there are no common factors left in the expansions. Both lists also include the number 3, so on a new line we can write 2 And 3 . After that, compare the expansions again: 30 = 2 x 3 x 5 and 36 = 2 x 2 x 3 x 3. As you can see, there are no common factors left in them.
      • To find the greatest common divisor, you must find the product of all common factors. In our example, these are 2 and 3, so gcd is 2 x 3 = 6 . This is the largest number by which the numbers 30 and 36 are divisible without a remainder.

   2. GCD can be used to simplify fractions. If you suspect that a fraction can be reduced, use the greatest common divisor. Use the above procedure to find the GCD of the numerator and denominator. Then divide the numerator and denominator of the fraction by that number. As a result, you will get the same fraction in a simpler form.

      • For example, let's simplify the fraction 30/36. As we stated above, for 30 and 36 GCD is 6, so we divide the numerator and denominator by 6:
      • 30 ÷ 6 = 5
      • 36 ÷ 6 = 6
      • 30 / 36 = 5 / 6

   3. Find the least common multiple of two numbers. The least common multiple (LCM) of two numbers is the smallest number that is evenly divisible by both given numbers. For example, the LCM of 2 and 3 is 6 because it is the smallest number that is divisible by 2 and 3. The following is an example of finding the LCM using prime factorization:

      • We start with two factorizations into prime factors. For example, for the number 126, the expansion can be written as 2 x 3 x 3 x 7. The number 84 is decomposed into prime factors in the form 2 x 2 x 3 x 7.
      • Let's compare how many times each factor occurs in expansions. Choose the list where the multiplier occurs the maximum number of times, and circle this place. For example, the number 2 occurs once in the expansion for 126 and twice in the list for 84, so circle 2x2 in the second multiplier list.
      • Repeat this action for each multiplier. For example, 3 occurs more often in the first expansion, so circle it 3x3. The number 7 appears once in both lists, so circle 7 (it doesn't matter in which list, if the given factor occurs in both lists the same number of times).
      • To find the LCM, multiply all circled numbers. In our example, the least common multiple of 126 and 84 is 2 x 2 x 3 x 3 x 7 = 252. This is the smallest number that is divisible by 126 and 84 without a remainder.

   4. Use LCM to add fractions. When adding two fractions, you need to bring them to a common denominator. To do this, find the LCM of two denominators. Then multiply the numerator and denominator of each fraction by such a number that the denominators of the fractions become equal to the LCM. After that, you can add fractions.

      • For example, you need to find the sum of 1 / 6 + 4 / 21 .
      • Using the above method, you can find the LCM for 6 and 21. It is equal to 42.
      • Let's transform the fraction 1/6 so that its denominator is 42. To do this, divide 42 by 6: 42 ÷ 6 = 7. Now multiply the numerator and denominator of the fraction by 7: 1/6 x 7/7 = 7/42.
      • To bring the second fraction to the denominator 42, divide 42 by 21: 42 ÷ 21 = 2. Multiply the numerator and denominator of the fraction by 2: 4 / 21 x 2 / 2 = 8 / 42.
      • After the fractions are reduced to the same denominator, they can be easily added: 7/42 + 8/42 = 15/42.

In this article you will find all the necessary information that answers the question, how to factorize a number. First, a general idea of ​​\u200b\u200bthe decomposition of a number into prime factors is given, examples of expansions are given. The canonical form of factoring a number into prime factors is shown next. After that, an algorithm for decomposing arbitrary numbers into prime factors is given, and examples of decomposing numbers using this algorithm are given. Alternative methods are also considered that allow you to quickly decompose small integers into prime factors using divisibility criteria and the multiplication table.

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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 the product of some numbers takes place, and the clarifying word “prime” 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 the given number must be represented 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 , so the factorization 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 . Separately, we emphasize that prime factors in the expansion can be repeated. This is clearly illustrated by the following example: 144=2 2 2 2 3 3 . But the representation of the form 45=3 15 is not a decomposition into prime factors, since the number 15 is composite.

The following question arises: “And what numbers can be decomposed 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. Given this fact and , it can be argued that the product of several prime factors is a positive integer greater than one. Therefore, factorization takes place only for positive integers that are greater than 1.

But do all integers greater than one factor into prime factors?

It is clear that there is no way to decompose simple integers into prime factors. This is because prime numbers have only two positive divisors, one and itself, so they cannot be represented as a product of two or more primes. If an integer z could be represented as a 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, it is believed that any prime number is itself its decomposition.

What about composite numbers? Do composite numbers decompose into prime factors, and are all composite numbers subject to such a decomposition? An affirmative answer to a number of these questions is given by the fundamental theorem of arithmetic. The fundamental theorem of arithmetic states that any integer a that is greater than 1 can be decomposed into a product of prime factors p 1 , p 2 , ..., p n , while the expansion has the form a=p 1 p 2 ... p n , and this the decomposition is unique, if we do not take into account the order of the factors

Canonical decomposition 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 the prime factor p 1 occur s 1 times in the decomposition of the number a, 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 writing 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 form is 609 840=2 4 3 2 5 7 11 2.

The canonical decomposition 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 decomposing a number into prime factors

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

The essence of the process of expansion of a positive integer and greater than one number a is clear from the proof of the main theorem of arithmetic. The meaning is to sequentially find the smallest prime divisors p 1 , p 2 , …,p n numbers a, a 1 , a 2 , …, a n-1 , which allows you to get 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 a n =1 is obtained, then the equality a=p 1 ·p 2 ·…·p n will give us the required decomposition of the number a into prime factors. Here it should also be noted that p 1 ≤p 2 ≤p 3 ≤…≤p n.

It remains to deal with finding the smallest prime divisors at each step, and we will have an algorithm for decomposing a number into prime factors. The prime number table will help us find prime divisors. Let's show how to use it to get 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 divisible is 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 a prime number, then its smallest prime divisor does not exceed the number , where - 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).

For example, let's show how to find the smallest prime divisor of the number 87. We take the number 2. We divide 87 by 2, we get 87:2=43 (remaining 1) (if necessary, see the 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 table of prime numbers, this is the number 3 . We divide 87 by 3, we get 87:3=29. So 87 is evenly divisible by 3, so 3 is the smallest prime divisor of 87.

Note that in the general case, in order to factorize the number a, we need a table of prime numbers up to a number no 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, we will 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 algorithm for factoring a number into prime factors. The algorithm for expanding 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 is itself its decomposition into prime factors. If a 1 is equal to 1, then we have a=p 1 ·a 1 and go to the next step.
• We find the smallest prime divisor p 2 of the number a 1 , for this we sequentially sort through the numbers from the table of prime numbers, starting with p 1 , after which we calculate a 2 =a 1:p 2 . If a 2 =1, then the desired decomposition of the number a into prime factors has the form a=p 1 ·p 2 . If a 2 is equal to 1, then we have a=p 1 ·p 2 ·a 2 and go to the next step.
• Going through the numbers from the table of primes, 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 desired decomposition of the number a into prime factors has the form a=p 1 ·p 2 ·p 3 . If a 3 is equal to 1, then we have a=p 1 ·p 2 ·p 3 ·a 3 and go to the next step.
• Find the smallest prime divisor p n of the number a n-1 by sorting through the primes, 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 required decomposition of the number a into prime factors: a=p 1 ·p 2 ·…·p n .

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

It remains only to consider a few examples of applying the obtained algorithm to decomposing numbers into prime factors.

Prime factorization examples

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

Example.

Factor the number 78 into prime factors.

Solution.

We start searching for the first smallest prime divisor p 1 of the number a=78 . To do this, we begin to sequentially sort through the prime numbers from the table of prime numbers. We take the number 2 and divide by it 78, we get 78:2=39. The number 78 was divided by 2 without a remainder, so p 1 \u003d 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 go 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 start enumeration of numbers from the table of primes, starting with p 1 =2 . Divide 39 by 2, we get 39:2=19 (remaining 1). Since 39 is not evenly divisible by 2, 2 is not its divisor. Then we take the next number from the table of prime numbers (the number 3) and divide by it 39, we get 39:3=13. Therefore, p 2 \u003d 3 is the smallest prime divisor of the number 39, while a 2 \u003d a 1: p 2 \u003d 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 go 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 (res. 6) and 13:11=1 (res. 2) . The next prime number is 13, and 13 is divisible by it without a remainder, therefore, the smallest prime divisor p 3 of the number 13 is the number 13 itself, and a 3 =a 2:p 3 =13:13=1. Since a 3 =1 , then this step of the algorithm is the last one, 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 factoring a number into prime factors, we find p 1 =2 and a 1 =a:p 1 =83 006:2=41 503 , whence 83 006=2 41 503 .

At the second step, we find out that 2 , 3 and 5 are not prime divisors of the number a 1 =41 503 , and 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 a 2 =5 929 is 7 , since 5 929:7=847 . Thus, p 3 =7 , a 3 =a 2:p 3 =5 929:7=847 , whence 83 006=2 7 7 847 .

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

The result obtained can be written as a 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 has no prime divisor that does not exceed ( 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. We give examples for clarification.

For example, we need to decompose the number 10 into prime factors. We know from the multiplication table that 2 5=10 , and the numbers 2 and 5 are obviously prime, so the prime factorization of 10 is 10=2 5 .

Another example. Using the multiplication table, we decompose the number 48 into prime factors. We know that six eight is 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 twice two is four, then we get the desired decomposition into prime factors 48=2 3 2 2 2 . Let's write this decomposition in the canonical form: 48=2 4 ·3 .

But when decomposing the number 3400 into prime factors, you can use the signs of divisibility. The signs of divisibility by 10, 100 allow us to assert that 3400 is divisible by 100, while 3400=34 100, and 100 is divisible by 10, while 100=10 10, therefore, 3400=34 10 10. And on the basis of the sign of divisibility by 2, it can be argued 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 required one. It remains only to rearrange the factors so that they go in ascending order: 3 400=2 2 2 5 5 17 . We 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 represent the number 75 as a product of prime factors. The sign of divisibility by 5 allows us to assert that 75 is divisible by 5, while we get that 75=5 15. And from the multiplication table we know that 15=3 5 , therefore, 75=5 3 5 . This is the desired decomposition of the number 75 into prime factors.

Bibliography.

• Vilenkin N.Ya. etc. Mathematics. Grade 6: textbook for educational institutions.
• Vinogradov I.M. Fundamentals of number theory.
• Mikhelovich Sh.Kh. Number theory.
• Kulikov L.Ya. and others. Collection of problems in algebra and number theory: Textbook for students of fiz.-mat. specialties of pedagogical institutes.

Consider the following expressions with powers of (a + b) n , where a + b is any binomial and n is an integer.

Each expression is a polynomial. In all expressions, you can notice features.

1. In each expression, there is one term more than the exponent n.

2. In each term, the sum of the powers is equal to n, i.e. the power to which the binomial is raised.

3. The powers start from the binomial power n and decrease towards 0. The last term has no factor a. The first term has no factor b, i.e. the powers of b start at 0 and increase to n.

4. Coefficients start at 1 and increase by certain values ​​up to "half way" and then decrease by the same values ​​back to 1.

Let's take a closer look at the coefficients. Suppose we want to find the value (a + b) 6 . According to the feature we just noticed, there should be 7 members here
a 6 + c 1 a 5 b + c 2 a 4 b 2 + c 3 a 3 b 3 + c 4 a 2 b 4 + c 5 ab 5 + b 6 .
But how can we determine the value of each coefficient, c i ? We can do this in two ways. The first method involves writing the coefficients in a triangle, as shown below. This is known as Pascal's triangle :


There are many features in the triangle. Find as many as you can.
You may have found a way to write the next line of numbers using the numbers on the line above. Units are always located on the sides. Each remaining number is the sum of the two numbers above that number. Let's try to find the value of the expression (a + b) 6 by adding the following line using the features we found:

We see that in the last line

first and last number 1 ;
the second number is 1 + 5, or 6 ;
the third number is 5 + 10, or 15 ;
the fourth number is 10 + 10, or 20 ;
the fifth number is 10 + 5, or 15 ; And
the sixth number is 5 + 1, or 6 .

So the expression (a + b) 6 will be equal to
(a + b) 6 = 1 a 6 + 6 a 5 b + 15 a 4 b 2 + 20 a 3 b 3 + 15 a 2 b 4 + 6 ab5+ 1 b6.

In order to raise to the power (a + b) 8 , we complement two lines to Pascal's triangle:

Then
(a + b) 8 = a 8 + 8a 7 b + 28a 6 b 2 + 56a 5 b 3 + 70a 4 b 4 + 56a 3 b 5 + 28a 2 b 6 + 8ab 7 + b 8 .

We can generalize our results as follows.

Newton's binomial using Pascal's triangle

For any binomial a + b and any natural number n,
(a + b) n = c 0 a n b 0 + c 1 a n-1 b 1 + c 2 a n-2 b 2 + .... + c n-1 a 1 b n-1 + c n a 0 b n ,
where the numbers c 0 , c 1 , c 2 ,...., c n-1 , c n are taken from the (n + 1) series of Pascal's triangle.

Example 1 Raise to the power: (u - v) 5 .

Solution We have (a + b) n , where a = u, b = -v, and n = 5. We use the 6th row of Pascal's triangle:
1 5 10 10 5 1
Then we have
(u - v) 5 = 5 = 1 (u)5+ 5 (u) 4 (-v) 1 + 10 (u) 3 (-v) 2 + 10 (u) 2 (-v) 3 + 5 (u)(-v) 4 + 1 (-v) 5 = u 5 - 5u 4 v + 10u 3 v 2 - 10u 2 v 3 + 5uv 4 - v 5 .
Note that the signs of the terms fluctuate between + and -. When the power of -v is an odd number, the sign is -.

Example 2 Raise to the power: (2t + 3/t) 4 .

Solution We have (a + b) n , where a = 2t, b = 3/t, and n = 4. We use the 5th row of Pascal's triangle:
1 4 6 4 1
Then we have

Binomial decomposition using factorial values

Suppose we want to find the value (a + b) 11 . The downside to using Pascal's triangle is that we have to compute all the previous rows of the triangle to get the required row. The following method avoids this. It also allows you to find a specific line - say the 8th line - without calculating all other lines. This method is useful in calculations, statistics and it uses binomial coefficient notation .
We can formulate Newton's binomial as follows.

Binomial Newton using factorial notation

For any binomial (a + b) and any natural number n,
.

Newton's binomial can be proved by mathematical induction. She shows why binomial coefficient .

Example 3 Raise to the power: (x 2 - 2y) 5 .

Solution We have (a + b) n , where a = x 2 , b = -2y, and n = 5. Then, using Newton's binomial, we have


Finally, (x 2 - 2y) 5 = x 10 - 10x 8 y + 40x 6 y 2 - 80x 4 y 3 + 80x 2 y 4 - 35y 5 .

Example 4 Raise to the power: (2/x + 3√x ) 4 .

Solution We have (a + b) n , where a = 2/x, b = 3√x , and n = 4. Then, using Newton's binomial, we get


Finally (2/x + 3√x ) 4 = 16/x 4 + 96/x 5/2 + 216/x + 216x 1/2 + 81x 2 .

Finding a Specific Member

Suppose we want to determine one or another member of a term from an expression. The method we have developed will allow us to find this term without calculating all the rows of Pascal's triangle or all the previous coefficients.

Note that in Newton's binomial gives us the 1st term, gives us the 2nd term, gives us the 3rd term, and so on. This can be summarized as follows.

Finding (k + 1) term

(k + 1) term of expression (a + b) n is .

Example 5 Find the 5th term in the expression (2x - 5y) 6 .

Solution First, note that 5 = 4 + 1. Then k = 4, a = 2x, b = -5y, and n = 6. Then the 5th term of the expression will be

Example 6 Find the 8th term in the expression (3x - 2) 10 .

Solution First, note that 8 = 7 + 1. Then k = 7, a = 3x, b = -2 and n = 10. Then the 8th term of the expression will be

Total number of subsets

Suppose the set has n objects. The number of subsets containing k elements is . The total number of subsets of a set is the number of subsets with 0 elements, as well as the number of subsets with 1 element, as well as the number of subsets with 2 elements, and so on. The total number of subsets of a set with n elements is
.
Now let's consider exponentiation (1 + 1) n:

.
So. the total number of subsets is (1 + 1) n , or 2 n . We have proved the following.

Total number of subsets

The total number of subsets of a set with n elements is 2 n .

Example 7 How many subsets does the set (A, B, C, D, E) have?

Solution The set has 5 elements, then the number of subsets is 2 5 , or 32.

Example 8 Wendy's restaurant chain offers the following toppings for hamburgers:
{ketchup, mustard, mayonnaise, tomatoes, lettuce, onion, mushrooms, olives, cheese}.
How many different types of hamburgers can Wendy offer, excluding hamburger sizes or quantities?

Solution The toppings for each hamburger are elements of a subset of the set of all possible toppings, and the empty set is just a hamburger. The total number of possible hamburgers will be

. Thus, Wendy can offer 512 different hamburgers.

If the function f(x) has derivatives of all orders on some interval containing the point a, then the Taylor formula can be applied to it:
,
Where rn- the so-called residual term or the remainder of the series, it can be estimated using the Lagrange formula:
, where the number x lies between x and a.

Function entry rules:

If for some value X rn→0 at n→∞, then in the limit the Taylor formula turns for this value into the convergent Taylor series:
,
Thus, the function f(x) can be expanded into a Taylor series at the considered point x if:
1) it has derivatives of all orders;
2) the constructed series converges at this point.

For a = 0 we get a series called near Maclaurin:
,
Expansion of the simplest (elementary) functions in the Maclaurin series:
exponential functions
, R=∞
Trigonometric functions
, R=∞
, R=∞
, (-π/2< x < π/2), R=π/2
The function actgx does not expand in powers of x, because ctg0=∞
Hyperbolic functions


Logarithmic functions
, -1
Binomial series
.

Example #1. Expand the function into a power series f(x)= 2x.
Solution. Let us find the values ​​of the function and its derivatives at X=0
f(x) = 2x, f( 0) = 2 0 =1;
f"(x) = 2x ln2, f"( 0) = 2 0 ln2=ln2;
f""(x) = 2x ln 2 2, f""( 0) = 2 0 log 2 2= log 2 2;
…
f(n)(x) = 2x ln n 2, f(n)( 0) = 2 0 ln n 2=ln n 2.
Substituting the obtained values ​​of the derivatives into the Taylor series formula, we get:

The radius of convergence of this series is equal to infinity, so this expansion is valid for -∞<x<+∞.

Example #2. Write a Taylor series in powers ( X+4) for the function f(x)= e x.
Solution. Finding the derivatives of the function e x and their values ​​at the point X=-4.
f(x)= e x, f(-4) = e -4 ;
f"(x)= e x, f"(-4) = e -4 ;
f""(x)= e x, f""(-4) = e -4 ;
…
f(n)(x)= e x, f(n)( -4) = e -4 .
Therefore, the desired Taylor series of the function has the form:

This expansion is also valid for -∞<x<+∞.

Example #3. Expand function f(x)=ln x in a series by degrees ( X- 1),
(i.e. in a Taylor series in the vicinity of the point X=1).
Solution. We find the derivatives of this function.
f(x)=lnx , , , ,

f(1)=ln1=0, f"(1)=1, f""(1)=-1, f"""(1)=1*2,..., f(n)=(- 1) n-1 (n-1)!
Substituting these values ​​into the formula, we get the desired Taylor series:

With the help of d'Alembert's test, one can verify that the series converges at ½x-1½<1 . Действительно,

The series converges if ½ X- 1½<1, т.е. при 0<x<2. При X=2 we obtain an alternating series that satisfies the conditions of the Leibniz test. For x=0 the function is not defined. Thus, the region of convergence of the Taylor series is the half-open interval (0;2].

Example #4. Expand the function in a power series.
Solution. In decomposition (1) we replace x by -x 2, we get:
, -∞

Example number 5. Expand the function in a Maclaurin series .
Solution. We have
Using formula (4), we can write:

substituting instead of x in the formula -x, we get:

From here we find: ln(1+x)-ln(1-x) = -
Expanding the brackets, rearranging the terms of the series and making a reduction of similar terms, we get
. This series converges in the interval (-1;1) since it is obtained from two series, each of which converges in this interval.

Comment .
Formulas (1)-(5) can also be used to expand the corresponding functions in a Taylor series, i.e. for the expansion of functions in positive integer powers ( Ha). To do this, it is necessary to perform such identical transformations on a given function in order to obtain one of the functions (1) - (5), in which instead of X costs k( Ha) m , where k is a constant number, m is a positive integer. It is often convenient to change the variable t=Ha and expand the resulting function with respect to t in the Maclaurin series.

This method is based on the theorem on the uniqueness of the expansion of a function in a power series. The essence of this theorem is that in the vicinity of the same point, two different power series cannot be obtained that would converge to the same function, no matter how its expansion is performed.

Example No. 5a. Expand the function in a Maclaurin series, indicate the area of ​​convergence.
Solution. First we find 1-x-6x 2 =(1-3x)(1+2x) , .
to elementary:

The fraction 3/(1-3x) can be viewed as the sum of an infinitely decreasing geometric progression with a denominator of 3x if |3x|< 1. Аналогично, дробь 2/(1+2x) как сумму бесконечно убывающей геометрической прогрессии знаменателем -2x, если |-2x| < 1. В результате получим разложение в степенной ряд

with convergence region |x|< 1/3.

Example number 6. Expand the function in a Taylor series in the vicinity of the point x = 3.
Solution. This problem can be solved, as before, using the definition of the Taylor series, for which it is necessary to find the derivatives of the functions and their values ​​at X=3. However, it will be easier to use the existing decomposition (5):
=
The resulting series converges at or -3

Example number 7. Write a Taylor series in powers (x -1) of the function ln(x+2) .
Solution.


The series converges at , or -2< x < 5.

Example number 8. Expand the function f(x)=sin(πx/4) in a Taylor series around the point x =2.
Solution. Let's make the replacement t=x-2:

Using expansion (3), in which we substitute π / 4 t for x, we get:

The resulting series converges to the given function at -∞< π / 4 t<+∞, т.е. при (-∞Thus,
, (-∞

Approximate calculations using power series

Power series are widely used in approximate calculations. With their help, with a given accuracy, you can calculate the values ​​of roots, trigonometric functions, logarithms of numbers, definite integrals. Series are also used in the integration of differential equations.
Consider the expansion of the function in a power series:

To calculate the approximate value of a function at a given point X, belonging to the region of convergence of the indicated series, the first n members ( n is a finite number), and the remaining terms are discarded:

To estimate the error of the obtained approximate value, it is necessary to estimate the discarded residual r n (x) . For this, the following methods are used:

• if the resulting series is character-alternating, then the following property is used: for an alternating series that satisfies the Leibniz conditions, the absolute value of the remainder of the series does not exceed the first discarded term.
• if the given series is of constant sign, then the series composed of the discarded terms is compared with an infinitely decreasing geometric progression.
• in the general case, to estimate the remainder of the Taylor series, you can use the Lagrange formula: a x ).

Example #1. Compute ln(3) to within 0.01.
Solution. Let's use the decomposition , where x=1/2 (see example 5 in the previous topic):

Let's check if we can discard the remainder after the first three terms of the expansion, for this we evaluate it using the sum of an infinitely decreasing geometric progression:

So we can discard this remainder and get

Example #2. Calculate to the nearest 0.0001.
Solution. Let's use the binomial series. Since 5 3 is the nearest integer cube to 130, it is advisable to represent the number 130 as 130=5 3 +5.



since the fourth term of the obtained sign-alternating series that satisfies the Leibniz test is already less than the required accuracy:
, so it and the terms following it can be discarded.
Many practically necessary definite or improper integrals cannot be calculated using the Newton-Leibniz formula, because its application is associated with finding an antiderivative, often not having an expression in elementary functions. It also happens that finding an antiderivative is possible, but unnecessarily laborious. However, if the integrand is expanded into a power series, and the integration limits belong to the interval of convergence of this series, then an approximate calculation of the integral with a predetermined accuracy is possible.

Example #3. Calculate the integral ∫ 0 1 4 sin (x) x to within 10 -5 .
Solution. The corresponding indefinite integral cannot be expressed in elementary functions, i.e. is an "impossible integral". The Newton-Leibniz formula cannot be applied here. Let us calculate the integral approximately.
Dividing term by term the series for sin x on x, we get:

Integrating this series term by term (this is possible, since the limits of integration belong to the interval of convergence of this series), we obtain:

Since the resulting series satisfies the conditions of Leibniz and it is enough to take the sum of the first two terms in order to obtain the desired value with a given accuracy.
Thus, we find
.

Example #4. Calculate the integral ∫ 0 1 4 e x 2 to within 0.001.
Solution.
. Let's check if we can discard the remainder after the second term of the resulting series.
0.0001<0.001. Следовательно, .