The use of the leibher's formula for derivatives. Derivatives of higher orders

The solution of applied tasks is reduced to the calculation of the integral, but it is not always possible to do exactly. Sometimes it is necessary to know the value of a specific integral with some degree of accuracy, for example, to a thousandth.

There are tasks when it should be found the approximate value of a specific integral with the necessary accuracy, then numerical integration is used as the method of simplicity, trapezoids, rectangles. Not all cases allow you to calculate it with a certain accuracy.

This article considers the application of the Newton-Labender formula. This is necessary to accurately calculate a specific integral. Detailed examples will be given, the replacement of the variable in a specific integral will be considered and find the values \u200b\u200bof a specific integral when integrating in parts.

Formula Newton Labitsa

Definition 1.

When the function y \u003d y (x) is continuous from the segment [A; b], and F (x) is one of the first functions of this segment, then formula Newton Labitsa Considered. We write it so ∫ a b f (x) d x \u003d f (b) - f (a).

This formula believes the main formula of integral calculus.

To produce proof of this formula, it is necessary to use the concept of an integral with an existing variable upper limit.

When the function y \u003d f (x) is continuous from the segment [a; b], then the value of the argument x ∈ A; b, and the integral has the form ∫ A x F (T) D T and is considered the function of the upper limit. It is necessary to adopt the designation of the function will take the form ∫ a x f (t) d t \u003d φ (x), it is continuous, and for it, the inequality of the form ∫ a x f (t) d t "\u003d φ" (x) \u003d f (x) is true.

We fix that the increment of the function φ (x) corresponds to the increment of the argument Δ x, it is necessary to use the fifth primary property of a specific integral and get

Φ (x + Δ x) - φ x \u003d ∫ Ax + Δ xf (t) dt - ∫ Axf (t) dt \u003d \u003d ∫ ax + δ xf (t) dt \u003d f (c) · x + δ x - x \u003d F (C) · Δ x

where is the value C ∈ X; X + Δ x.

Fix the equality in the form φ (x + δ x) - φ (x) δ x \u003d f (c). By definition of the derivative function, it is necessary to move to the limit at Δ x → 0, then we obtain the formula of the form φ "(x) \u003d f (x). We obtain that φ (x) is one of the primitive species for the function y \u003d f (x), Located on [a; b]. Otherwise, the expression can be recorded

F (x) \u003d φ (x) + c \u003d ∫ a x f (t) d t + c, where C value is constant.

Calculate F (A) using the first property of a specific integral. Then we get that

F (a) \u003d φ (a) + C \u003d ∫ A A F (T) D T + C \u003d 0 + C \u003d C, hence we obtain that C \u003d F (a). The result is applicable when calculating F (B) and get:

F (b) \u003d φ (b) + c \u003d ∫ ABF (T) DT + C \u003d ∫ ABF (T) DT + F (A), in other words, f (b) \u003d ∫ ABF (T) DT + F ( a). Equality proves the formula of Newton labnica ∫ a b f (x) d x + f (b) - f (a).

The increment of the function is adopted as f x a b \u003d f (b) - f (a). With the designation of the Newton-Leibnia formula, takes the form ∫ a b f (x) d x \u003d f x a b \u003d f (b) - f (a).

To apply the formula, it is necessary to know one of the primitive y \u003d f (x) of the integrand function y \u003d f (x) from the segment [a; b], Calculate the increment of primitive from this segment. Consider a slightly calculation example using the Newton-Labender formula.

Example 1.

Calculate a specific integral ∫ 1 3 x 2 D x according to the Newton-Labender formula.

Decision

Consider that the integrated function of the form y \u003d x 2 is continuous from the segment [1; 3], then integrate on this segment. According to the table of uncertain integrals, we see that the function y \u003d x 2 has many primitive for all valid values \u200b\u200bX, which means x ∈ 1; 3 will be recorded as f (x) \u003d ∫ x 2 d x \u003d x 3 3 + c. It is necessary to take a primitive with C \u003d 0, then we obtain that f (x) \u003d x 3 3.

We use the Newton labnic formula and we obtain that the calculation of a specific integral will take the form ∫ 1 3 x 2 D x \u003d x 3 3 1 3 \u003d 3 3 3 - 1 3 3 \u003d 26 3.

Answer: ∫ 1 3 x 2 D x \u003d 26 3

Example 2.

Calculate a specific integral ∫ - 1 2 x · e x 2 + 1 d x by Newton-leibice formula.

Decision

The specified function is continuous from the segment [- 1; 2], it means that it is integrable. It is necessary to find the value of an indefinite integral ∫ x · ex 2 + 1 dx using the method of submission to the differential sign, then we obtain ∫ x · ex 2 + 1 dx \u003d 1 2 ∫ ex 2 + 1 d (x 2 + 1) \u003d 1 2 2 + 1 + c.

From here we have many primitive functions y \u003d x · e x 2 + 1, which are valid for all x, x ∈ - 1; 2.

It is necessary to take a primitive when C \u003d 0 and apply the Newton-Labender formula. Then we get the expression

∫ - 1 2 x · ex 2 + 1 dx \u003d 1 2 EX 2 + 1 - 1 2 \u003d 1 2 E 2 2 + 1 - 1 2 E (- 1) 2 + 1 \u003d 1 2 E (- 1) 2 + 1 \u003d 1 2 E 2 (E 3 - 1)

Answer: ∫ - 1 2 x · e x 2 + 1 D x \u003d 1 2 E 2 (E 3 - 1)

Example 3.

Calculate the integrals ∫ - 4 - 1 2 4 x 3 + 2 x 2 D x and ∫ - 1 1 4 x 3 + 2 x 2 D x.

Decision

Cut - 4; - 1 2 suggests that the function under the integral sign is continuous, it means that it is integrated. From here we will find many primitive functions y \u003d 4 x 3 + 2 x 2. We get that

∫ 4 x 3 + 2 x 2 D x \u003d 4 ∫ x D x + 2 ∫ x - 2 d x \u003d 2 x 2 - 2 x + c

It is necessary to take a primitive f (x) \u003d 2 x 2 - 2 x, then by applying the Newton-Labender formula, we get an integral that calculates:

∫ - 4 - 1 2 4 x 3 + 2 x 2 dx \u003d 2 x 2 - 2 x - 4 - 1 2 \u003d 2 - 1 2 2 - 2 - 1 2 - 2 - 4 2 - 2 - 4 \u003d 1 2 + 4 - 32 - 1 2 \u003d - 28

We produce a transition to the calculation of the second integral.

From the segment [- 1; 1] we have that the integrated function is considered unlimited, because Lim X → 0 4 x 3 + 2 x 2 \u003d + ∞, then it follows from this that a necessary condition for integrability from the segment. Then f (x) \u003d 2 x 2 - 2 x is not primitive for Y \u003d 4 x 3 + 2 x 2 from the segment [- 1; 1], since the point O belongs to the segment, but is not included in the definition area. It means that there is a certain integral of Riemann and Newton leibher for the function y \u003d 4 x 3 + 2 x 2 from the segment [- 1; one ] .

Answer: ∫ - 4 - 1 2 4 x 3 + 2 x 2 D x \u003d - 28,there is a specific integral of Riemann and Newton labnice for the function y \u003d 4 x 3 + 2 x 2 of the segment [- 1; one ] .

Before using the Newton-Labitsa formula, you need to know exactly about the existence of a specific integral.

Replacing the variable in a specific integral

When the function y \u003d f (x) is defined and continuous from the segment [A; b], then the existing set [A; b] is considered the area of \u200b\u200bthe values \u200b\u200bof the function x \u003d g (z) defined on the segment α; β with an existing continuous derivative, where G (α) \u003d a and g β \u003d B, we obtain that ∫ a b f (x) d x \u003d ∫ α β f (g (z)) · g "(z) d z.

This formula is used when it is necessary to calculate the integral ∫ a b f (x) D x, where the indefinite integral has the form ∫ f (x) d x, calculate using the substitution method.

Example 4.

Calculate a specific integral of the form ∫ 9 18 1 x 2 x - 9 D x.

Decision

The integrand is considered a continuous integration on the intercom, which means a certain integral takes place on existence. We give the designation that 2 x - 9 \u003d z ⇒ x \u003d g (z) \u003d z 2 + 9 2. The value x \u003d 9, it means that z \u003d 2 · 9 - 9 \u003d 9 \u003d 3, and at x \u003d 18 we obtain that z \u003d 2 · 18 - 9 \u003d 27 \u003d 3 3, then g α \u003d g (3) \u003d 9, g β \u003d g 3 3 \u003d 18. When substituting the values \u200b\u200bobtained in the formula ∫ A b f (x) d x \u003d ∫ α β f (g (z)) · g "(z) d z, we get that

∫ 9 18 1 x 2 x - 9 dx \u003d ∫ 3 3 3 1 z 2 + 9 2 · z · z 2 + 9 2 "dz \u003d ∫ 3 3 3 1 z 2 + 9 2 · z · zdz \u003d ∫ 3 3 3 2 z 2 + 9 dz

According to the table of uncertain integrals, we have that one of the primitive functions 2 Z 2 + 9 takes the value 2 3 A R C T G Z 3. Then, when applying the Newton-Labitsa formula, we get that

∫ 3 3 3 2 z 2 + 9 D z \u003d 2 3 a r c t g z 3 3 3 3 \u003d 2 3 a r c t g 3 3 3 - 2 3 a r c t g 3 3 \u003d 2 3 a r c t g 3 - a r c t g 1 \u003d 2 3 π 3 - π 4 \u003d π 18

Finding could be made without using the formula ∫ A B f (x) d x \u003d ∫ α β f (g (z)) · g "(z) d z.

If, with the method of replacement, use the integral of the form ∫ 1 x 2 x - 9 D x, then you can come to the result ∫ 1 x 2 x - 9 D x \u003d 2 3 a r c T G 2 x - 9 3 + c.

From here, we will calculate the formula of Newton Labits and calculate a specific integral. We get that

∫ 9 18 2 Z 2 + 9 DZ \u003d 2 3 ARCTGZ 3 9 18 \u003d 2 3 3 ARCTG 2 · 18 - 9 3 - Arctg 2 · 9 - 9 3 \u003d 2 3 3 Arctg 3 - Arctg 1 \u003d 2 3 π 3 - π 4 \u003d π 18

The results coincided.

Answer: ∫ 9 18 2 x 2 x - 9 D x \u003d π 18

Integration in parts when calculating a specific integral

If on the segment [a; b] defined and continuous functions u (x) and v (x), then their first-order derivatives v "(x) · u (x) are integrable, thus from this segment for the integrable function U" (x) · V ( x) Equality ∫ ABV "(x) · U (x) dx \u003d (u (x) · v (x)) AB - ∫ ABU" (x) · V (x) DX is valid.

The formula can be used then, it is necessary to calculate the integral ∫ a b f (x) d x, and ∫ F (x) D x must be sought using integration in parts.

Example 5.

Calculate a specific integral ∫ - π 2 3 π 2 x · sin x 3 + π 6 D x.

Decision

The function x · sin x 3 + π 6 is integrable on the segment - π 2; 3 π 2, it means it is continuous.

Let U (x) \u003d x, then D (V (x)) \u003d V "(x) dx \u003d sin x 3 + π 6 dx, and D (u (x)) \u003d u" (x) dx \u003d dx, and V (x) \u003d - 3 cos π 3 + π 6. From the formula ∫ a b v "(x) · u (x) d x \u003d (u (x) · v (x)) a b - ∫ a b u" (x) · v (x) d x we \u200b\u200bget that

∫ - π 2 3 π 2 x · sin x 3 + π 6 dx \u003d - 3 x · cos x 3 + π 6 - π 2 3 π 2 - ∫ - π 2 3 π 2 - 3 cos x 3 + π 6 dx \u003d \u003d - 3 · 3 π 2 · cos π 2 + π 6 - - 3 · - π 2 · cos - π 6 + π 6 + 9 sin x 3 + π 6 - π 2 3 π 2 \u003d 9 π 4 - 3 π 2 + 9 sin π 2 + π 6 - sin - π 6 + π 6 \u003d 9 π 4 - 3 π 2 + 9 3 2 \u003d 3 π 4 + 9 3 2

The example solution can be performed in a different way.

Find many primitive functions x · sin x 3 + π 6 using integration in parts using Newton-Leibnia formula:

∫ x · sin xx 3 + π 6 dx \u003d u \u003d x, dv \u003d sin x 3 + π 6 dx ⇒ du \u003d dx, v \u003d - 3 cos x 3 + π 6 \u003d \u003d - 3 cos x 3 + π 6 + 3 ∫ cos x 3 + π 6 dx \u003d - 3 x cos x 3 + π 6 + 9 sin x 3 + π 6 + c ⇒ ∫ - π 2 3 π 2 x · sin x 3 + π 6 dx \u003d - 3 cos x 3 + π 6 + 9 sincos x 3 + π 6 - - - 3 · - π 2 · cos - π 6 + π 6 + 9 sin - π 6 + π 6 \u003d 9 π 4 + 9 3 2 - 3 π 2 - 0 \u003d 3 π 4 + 9 3 2

Answer: ∫ x · sin x x 3 + π 6 d x \u003d 3 π 4 + 9 3 2

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Derivatives of higher orders

In this lesson, we will learn to find derivatives of higher orders, as well as record the general formula of the Anna derivative. In addition, the Formula of the Leibnia will be considered and on numerous requests - derivatives of higher orders from implicitly specified function. I propose to immediately go through a mini test:

Here is a function: And here is its first derivative:

In the event that you have any difficulty / misunderstanding about this example, please start with two basic articles of my course: How to find a derivative? and Derivative complex function. After the development of elementary derivatives, I recommend to get acquainted with the lesson Simplest tasks with derivativewhere we figured out, in particular with the second derivative.

It is easy to even guess that the second derivative is a derivative of the 1st derivative:

In principle, the second derivative is already considered the derivative of the highest order.

Similarly: The third derivative is a derivative of the 2nd derivative:

The fourth derivative is derived from the 3rd derivative:

Fifth derivative: , and it is obvious that all the derivatives of higher orders will also be zero:

In addition to Roman numbering in practice, the following notation is often used:
The derivative of the "enyn" order is denoted by. At the same time, the tentary index must be configured in brackets - To distinguish a derivative from "Games" to the degree.

Sometimes there is such an entry: - Third, fourth, fifth, ..., "enons" derivatives, respectively.

Forward without fear and doubts:

Example 1.

Dana feature. To find .

Decision: What do you hurt ... - Forward for the fourth derivative :)

The four strokes are no longer accepted, so we go to numeric indices:

Answer:

Well, and now think about such a question: what to do if, by condition, it is necessary to find not 4th, but for example, the 20th derivative? If for the 3-4-5-year derivative (maximum, 6-7th) The procedure for the decision is made quite quickly, then before the derivatives of higher orders we will "DOES" Oh, how soon. Do not write, in fact, 20 lines! In such a situation, we need to analyze several different derivatives, to see the pattern and make the formula of the Annna derivative. So, in example number 1 it is easy to understand that with each next differentiation before the exponent will "pop up" an additional "Troika" will "pop up", and at any step, the degree of "Troika" is equal to the derivative number, therefore:

Where - an arbitrary natural number.

And indeed, if, it turns out exactly the 1st Derivative: If - then the 2nd: etc. Thus, the twentieth derivative is determined instantly: - and no "kilometer sheets"!

We warm yourself:

Example 2.

Find functions. Place a derivative order

Solution and answer at the end of the lesson.

After an invigorating workout, we consider more complex examples in which we will work out the above solution algorithm. Those who managed to familiarize themselves with the lesson Sequence limitwill be slightly easier:

Example 3.

Find for a function.

Decision: To clarify the situation will find several derivatives:

The received numbers multiply not in a hurry! ;-)

Perhaps enough. ... even moved a bit.

In the next step, it is best to make a formula "Anna" derivative (since soon, the condition does not require this, then you can do the Chernivik). For this, we look at the results and reveal the patterns that each next derivative is obtained.

First, they are alternate. Alignment provides "Flashing"And since the 1st derivative is positive, the following multiplier will enter the general formula: . It is suitable and equivalent to the option, but I personally, as an optimist, love the sign "Plus" \u003d)

Secondly, in the numerator "winding" factorialAnd he "lags behind" from the number of the derivative per unit:

And thirdly, the degree of "two" is growing in the numerator, which is equal to the derivative number. The same can be said about the degree of denominator. Finally:

For test purposes, we will substitute a couple of "En" values, for example, and:

Wonderful, now allow a mistake - just a sin:

Answer:

A more simple feature for self solutions:

Example 4.

Find functions.

And the task is growing:

Example 5.

Find functions.

Once again we repeat the procedure:

1) First we find several derivatives. To catch patterns usually grabs three or four.

2) then strongly recommend making (at least on the draft) "Anna" derivative - it will be guaranteed to save the mistakes. But you can do without, i.e. To mentally estimate and immediately burn, for example, the twentieth or eighth derivative. Moreover, some people are generally able to solve the tasks in question orally. However, it should be remembered that the "fast" methods are fraught, and it is better to be restrained.

3) At the final stage, perform an inspection of the "enna" derivative - take a couple of "En" values \u200b\u200b(better adjacent) and perform substitution. And even more reliable - check all the previously found derivatives. After that, we substitute in the desired value, for example, or and neatly having the result.

Summary of 4 and 5 examples at the end of the lesson.

In some tasks, in order to avoid problems, you need to sit over the function:

Example 6.

Decision: Differentiate the proposed function does not want at all, because it turns out a "bad" fraction that will greatly find the following derivatives.

In this regard, it is advisable to perform preliminary transformations: use square difference formula and property logarithm :

Other things:

And old girlfriends:

I think everything is visible. Please note that the 2nd fraction is alternated, and the 1st - no. Construct a derivative order:

Control:

Well, for beauty, I will bring factorial for brackets:

Answer:

Interesting task for self solutions:

Example 7.

Write a procedure for a derivative order for a function

And now about the unshakable circular order, which even the Italian mafia will envy:

Example 8.

Dana feature. To find

Eighteenth derivative at point. Just.

Decision: First, obviously, you need to find. Go:

Sinus began to sinus and came. It is clear that with further differentiation, this cycle will continue indefinitely, and the following question arises: how best to "get" to the eighteenth derivative?

Method "Amateur": Quickly recording the right in the number of subsequent derivatives:

In this way:

But it works if the order of the derivative is not too large. If you need to find, say, a hundred derivative, then you should use a division by 4. One hundred is divided into 4 without a residue, and it is easy to see that such numbers are located at the bottom line, so :.

By the way, the 18th derivative can also be determined from similar considerations:
In the second line there are numbers that are divided by 4 with the residue 2.

Another, more academic method is based on the frequency of sinus and formulas of the cast. We use the finished formula of the "enna" derivative of sine In which the desired number is simply substituted. For example:
(the formula of the cast ) ;
(the formula of the cast )

In our case:

(1) Since sinus is a periodic function with a period, then the argument can be painlessly "unscrew" 4 periods (i.e.).

The derivative of the order from the work of two functions can be found by the formula:

In particular:

Specially remember nothing, because, the more formulas you know - the less you understand. It is much more useful to get acquainted with binom NewtonSince the Formula Leibnia is very similar to him. Well, those lucky who will get a derivative of the 7th or higher orders (What, however, unlikely)will be forced to do it. However, when the cherodes will reach combinatorics - it will still have \u003d)

Find the third derivative function. We use the Formula Leibnitsa:

In this case: . Derivatives easy to overdo it:

Now neatly and carefully carry out the substitution and simplify the result:

Answer:

Derivatives of higher orders from the functions defined implicitly

Many of us spent long hours, days and weeks of life to study circles, parabol, hyperbol - And sometimes it generally seemed to be a punishment. So let's avenge and fit them as follows!

Let's start with the "school" parabola in her canonical position:

Example 14.

An equation is given. To find .

Decision: The first step is well acquainted:

The fact that the function and its derivative are implicitly implicit the essence of the case does not change, the second derivative is a derivative of the 1st derivative:

However, there are their rules of the game: the derivatives of the 2nd and higher orders are accepted to express only through "X" and "Igarek". Therefore, in the resulting 2-derivative substitution:

The third derivative is derived from the 2nd derivative:

Similarly, we will substitute:

Answer:

"School" hyperbole in canonical position - For independent work:

Example 15.

An equation is given. To find .

I repeat that the 2nd derivative and the result should be expressed only through "X" / "IKRAR"!

A brief solution and answer at the end of the lesson.

After children's pranks, we will look at the German subparthrix @ FIU Consider more adult examples from which we learn another important technique of solutions:

Example 16.

Ellipse himself.

Decision: Find the 1st Derivative:

And now we will stop and analyze the next moment: the fraction is to be differentiated that it is not happy at all. In this case, it, of course, is simple, but in the actual tasks of such gifts two times and turned around. Is there a way to avoid finding a bulky derivative? Exists! We take the equation and use the same reception as when I find the 1st Derivative - "hang" strokes on both parts:

The second derivative must be expressed only through and, so now (right now) It is convenient to get rid of the 1st derivative. To do this, we substitute for the resulting equation:

To avoid unnecessary technical difficulties, multiply both parts on:

And only at the final stage we decorated the fraction:

Now we look at the initial equation and notice that the result obtained is simplified:

Answer:

How to find a value of the 2nd derivative at any point (which, it is clear, belongs to the ellipse), for example, at the point ? Very easy! This motive has already met at the lesson about equation Normal: In the expression of the 2nd derivative you need to substitute :

Of course, in all three cases it is possible to obtain explicitly specified functions and differentiate them, but then morally tune in to work with two functions that contain roots. In my opinion, the solution is more convenient to conduct "implicit way".

Final example for self solutions:

Example 17.

Find an implicitly specified function

A leibher formula is given to calculate the N-th derivative of the product of two functions. It is given its proof in two ways. An example of calculating the N-order derivative is considered.

Content

See also: Derivative work of two functions

Formula Leibnitsa

Using the laboratory formula, you can calculate the nth order derivative from the product of two functions. It has the following form:
(1) ,
Where
- Binomial coefficients.

Binomial coefficients are coefficients of decomposition of binoma in degrees and:
.
Also, the number is the number of combinations from N by k.

Proof of Formula Leibniz

We apply the formula for the derivative of the work of two functions:
(2) .
We rewrite formula (2) in the following form:
.
That is, we believe that one function depends on the variable x, and the other - from the variable y. At the end of the calculation, we assume. Then the previous formula can be written as:
(3) .
Since the derivative is equal to the amount of members, and each member is a product of two functions, then to calculate the derivatives of the highest order, can be consistently applied to the rule (3).

Then for the n-order derivative we have:

.
Considering that we, we get the Formula Leibnitsa:
(1) .

Proof by induction

We present proof of the formula of the leibher by the method of mathematical induction.

Once again, repel the Formula Leibnitsa:
(4) .
For n \u003d 1, we have:
.
This is a formula for a derivative product of two functions. She is fair.

Suppose that formula (4) is valid for the derivative of the N-order. We prove that it is valid for the derivative N + 1 -o order.

Differentiasis (4):
;

.
So, we found:
(5) .

Substitute in (5) and consider::

.
It can be seen that formula (4) has the same appearance for the derivative N + 1 -o order.

So, formula (4) is valid for n \u003d 1 . From the assumption that it is performed, for some number N \u003d M it follows that it is performed for n \u003d m + 1 .
Formula Leibnitsa is proved.

Example

Calculate the N-th derivative function
.

Apply Formula Leibnica
(2) .
In our case
;
.

On the table derivatives we have:
.
Apply the properties of trigonometric functions:
.
Then
.
It can be seen that the differentiation of the sine function leads to its shift. Then
.

We find derivatives from the function.
;
;
;
, .

Since when, in the formula, the leibher is different from zero only the first three members. We find binomial coefficients.
;
.

By the Formula, Leibnia have:

.

See also:

The text of the work is placed without images and formulas.
The full version of the work is available in the "Work Files" tab in PDF format

"To me too, Binin Newton!»

from the novel "Master and Margarita"

"The triangle of Pascal is so simple that even a ten-year-old child can write it. At the same time, he pays inexhaustible treasures and binds together various aspects of mathematics that have nothing to do at first glance. Such unusual properties make it possible to consider the triangle of Pascal one of the most elegant schemes in the whole mathematics "

Martin Gardner.

Purpose of work: To summarize the formula of abbreviated multiplication, show their application to solving problems.

Tasks:

1) to explore and systematize information on this issue;

2) Disassemble examples of tasks for the use of Newton Binoma and the formulas of the amount and difference of degrees.

Research objects: Binin Newton, the formula of the amount and difference of degrees.

Research methods:

Work with educational and popular literature, Internet resources.

Calculations, comparison, analysis, analogy.

Relevance.A person often has to deal with the tasks in which you need to calculate the number of all possible ways of location of some items or the number of all possible methods for implementing some action. Different paths or options that you have to choose a person are folded into a wide variety of combinations. And a whole section of mathematics, called the combinatorics, is engaged in searching for answers to questions: how many combinations are in one case or another.

Combinators have to deal with many specialties: chemistry scientist, biologist, constructor, dispatcher, etc. Strengthening interest in the combinatorics recently is caused by the rapid development of cybernetics and computing equipment.

Introduction

When they want to emphasize that the interlocutor exaggerates the complexity of the tasks with whom he encountered, they say: "To me, Binin Newton!" Say, here Bin Nudon, it is difficult, and what problems do you have! Not even those people who whose interests are not connected with mathematics were heard about Newton's Biinoma.

The word "bin" means biccoon, i.e. The sum of the two terms. From the school year, the so-called formulas of abbreviated multiplication are known:

( but + b) 2 \u003d A. 2 + 2ab + B 2 , (A + B) 3 \u003d A. 3 + 3A 2 b + 3AB 2 + B. 3 .

The generalization of these formulas is a formula called Newton's Binomine formula. Used in school and formulas decomposition on multipliers of square differences, amounts and differences of cubes. Do they have a generalization for other degrees? Yes, there are such formulas, they are often used in solving various tasks: on evidence of divisibility, reduction of fractions, approximate calculations.

The study of generalizing formulas is developing deductive-mathematical thinking and general thinking abilities.

Section 1. Newton Binoma Formula

Combinations and their properties

Let X be a set consisting of n elements. Any subset of Y of the set x, containing k elements, is called the combination of K elements from N, while, k ≤ n.

The number of different combinations of K elements from N is denoted by n k. One of the most important formulas for combinatorics is the following formula for the number with N K:

It can be recorded after obvious contractions as follows:

In particular,

This is quite consistent with the fact that in the set x there is only one subset of 0 elements - an empty subset.

The numbers C n K have a number of wonderful properties.

Formula is valid with n k \u003d with n - k n, (3)

The meaning of formula (3) is that there is a mutually unequivocal correspondence between the plurality of all K-membrane subsets from X and the set of all (n - k) -thded subsets from X: To establish this correspondence, it is sufficient to each K-membered subset Y Match its addition in the set X.

Formula C 0 n + C 1 n + C 2 n + ... + with n n \u003d 2 n (4)

The amount that stands in the left side expresses the number of all subsets of the set x (C 0 n is the number of 0-membered subsets, C 1 n is the number of single subsets, etc.).

With any k, 1≤ k≤ n, equality is fair

C k n \u003d C n -1 k + C n -1 k -1 (5)

This equality is easy to obtain with the help of formula (1). Indeed,

1.2. The output of the Binoma Newton formula

Consider the degrees of bounce a +.b. .

n \u003d 0, (a +b. ) 0 = 1

n \u003d 1, (a +b. ) 1 \u003d 1A + 1b.

n \u003d 2,(A +.b. ) 2 \u003d 1A. 2 + 2A.b. +1 b. 2

n \u003d 3,(A +.b. ) 3 \u003d 1 A. 3 + 3A. 2 b. + 3A.b. 2 +1 b. 3

n \u003d 4,(A +.b. ) 4 \u003d 1A. 4 + 4A. 3 b. + 6A. 2 b. 2 + 4A.b. 3 +1 b. 4

n \u003d 5,(A +.b. ) 5 = 1A. 5 + 5A. 4 b. + 10A. 3 b. 2 + 10A. 2 b. 3 + 5A.b. 4 + 1 b. 5

Note the following monomerities:

The number of members of the resulting polynomial per unit is greater than the indicator of the degree of binoma;

The indicator of the degree of the first term decreases from n to 0, the indicator of the degree of the second term increases from 0 to n;

The degrees of all single-panels are equal to the degree of bounced in the condition;

Each single-wing is the product of the first and second expression in various degrees and some number - the binomine coefficient;

Binominal coefficients equal to the beginning and end of the decomposition are equal.

The generalization of these formulas is the following formula called Newton's Binomine formula:

(a. + b. ) n. = C. 0 n. a. n. b. 0 + C. 1 n. a. n. -1 b. + C. 2 n. a. n. -2 b. 2 + ... + C. n. -1 n. aB n. -1 + C. n. n. a. 0 b. n. . (6)

In this formula n. Maybe any natural number.

We derive formula (6). First of all, we write:

(a. + b. ) n. = (a. + b. )(a. + b. ) ... (a. + b. ), (7)

where the number of variable brackets is equal n.. From the usual rule of multiplication of the amount in the amount implies that the expression (7) is equal to the sum of all sorts of works, which can be as follows: anyone's first summary a + B. multiplied by anyone's second sum a + B., anyone's third sum, etc.

From what is clear that the term in expression for (a. + b. ) n. correspond (mutually unambiguous) strings of n, composed of letters a and b. Among the components will meet such members; Obviously, such members correspond to strings containing the same number of letters. but. But the number of lines containing exactly k times letter butEqually with n k. It means that the sum of all members containing the letter and the multiplier is exactly k times, equal to n k a. n. - k. b. k. . Since K can take values \u200b\u200b0, 1, 2, ..., n - 1, n, then formula (6) follows from our argument. Note that (6) you can record shorter: (8)

Although the formula (6) is called Newton's name, in reality she was revealed before Newton (for example, Pascal knew it). The merit of Newton is that he found a generalization of this formula in case not entire indicators. It is I.Nyuton in 1664-1665. He brought the formula that expresses twisted for arbitrary fractional and negative indicators.

The numbers C 0 n, C 1 n, ..., C n n, which are included in formula (6), are called binomial coefficients, which are determined as follows:

From formula (6), you can get a number of properties of these coefficients. For example, believed but \u003d 1, b \u003d 1, we get:

2 n \u003d C 0 n + C 1 n + C 2 n + C 3 n + ... + C n n,

those. formula (4). If put but \u003d 1, b \u003d -1, then we will have:

0 \u003d C 0 n - C 1 n + C 2 n - C 3 n + ... + (-1) n c n n

or with 0 n + C 2 n + C 4 n + ... \u003d C 1 n + C 3 n + + C 5 n + ....

This means that the sum of the coefficients of active decomposition members is equal to the sum of the coefficients of odd decomposition members; Each of them is 2 N -1.

The coefficients of members equidistant from the ends of the decomposition are equal. These properties follows from the relation: with n k \u003d with n n - k

Interesting a private case

(x + 1) n \u003d c 0 n x n + c 1 n x n-1 + ... + c k n x n - k + ... + c n n x 0

or in short (x +1) n \u003d σc n k x n - k.

1.3. Polynomial theorem.

Theorem.

Evidence.

So that after disclosure, the brackets turned out to be unrochene, you need to choose those brackets from which it is taken, those brackets from which is taken, etc. And those brackets from which are taken. The coefficient at the same time, after bringing such members, is equal to the number of ways that such a choice can be implemented. The first step of the election sequence can be carried out by methods, the second step -, the third -, etc., -Y step in the ways. The desired coefficient is equal to the work

Section 2. Derivatives of higher orders.

The concept of derivatives of higher orders.

Let the function differentiate in some interval. Then its derivative, generally speaking, depends on h.that is, is a function from h.. Consequently, in relation to it, it is possible to raise the question of the existence of the derivative.

Definition . The derivative of the first derivative is called a second-order derivative or a second derivative and is indicated by the symbol or, that is,

Definition . The derivative of the second derivative is called a third-order derivative or a third derivative and is indicated by the symbol or.

Definition . Derivativen. orderfunctions called the first derivative of the derivative (n. -1) -go order of this function and is indicated by the symbol or:

Definition . Derivatives of order above the first are called highest derivatives.

Comment. Similarly, you can get a formula n. The derivative function:

The second derivative of the parametrically specified function

If the function is set by parametric equations, it is necessary to use the expression for its first derivative to find the second order derivative, as a complex function of an independent variable.

Since, then

and taking into account the fact that

We get, that is.

Similarly, you can find the third derivative.

Differential sum, works and private.

Since the differential is obtained from an independent variable multiplication derivative, then, knowing the derivatives of the main elementary functions, as well as the rules for finding derivatives, one can come to the same rules for finding differentials.

1 0 . Differential constant is zero.

2 0 . The differential of the algebraic amount of the finite number of differentiable functions is equal to the algebraic amount of differentials of these functions .

3 0 . The differential of the work of two differentiable functions is equal to the amount of works of the first function to the differential of the second and second function on the differential of the first .

Corollary. A permanent multiplier can be taken out of the differential sign.

2.3. Functions specified parametrically, their differentiation.

Definition . The function is called a given parametric, if both variables h. and es are defined each individually as unambiguous functions from the same auxiliary variable - the parametert. :

wheret. varies within.

Comment . We present parametric circle and ellipse equations.

a) Circumfied with the center at the beginning of the coordinates and radius r. Has parametric equations:

b) We write parametric equations for the ellipse:

By excluding the parameter t. From the parametric equations of the lines under consideration, it is possible to come to their canonical equations.

Theorem . If the function u from argument x is given by parametric equations, where and differentiablet. Functions and, then.

2.4. Formula Leibniza

To find a derivative n. -Oh order from the work of two functions is a large practical value of the Labitsa formula.

Let be u. and v. - Some functions from variable h.having derivatives of any order and y. = uV . Express n. Derivatives through derived functions u. and v. .

We have sequential

It is easy to notice an analogy between expressions for the second and third derivatives and decomposition of the Newton Binoma, respectively, in the second and third degrees, but instead of the indicators of the degree cost the number that determine the procedure for the derivative, and the functions themselves can be considered as "zero-order derivatives". Given this, we get a Formula Leibnitsa:

This formula can be proved by mathematical induction.

Section 3. Application of the Labender Formula.

To calculate the derivative of any order from the product of two functions, bypassing the sequential use of the formula for calculating the derivative from the product of two functions, applies formula Leibniza.

Using this formula, consider examples of calculating the N-th order derivative from the product of two functions.

Example 1.

Find a second order derivative function

According to the definition, the second derivative is the first derivative of the first derivative, that is

Therefore, we first find a first-order derivative from a given function according to differentiation rules and Using table derivatives:

Now we find a derivative of the first order derivative. This will be the desired second order derivative:

Answer:

Example 2.

Find a derivative of the order of the function

Decision.

We will consistently find the derivatives of the first, second, third, and so on the order of the specified function in order to establish a pattern that can be generalized by a derivative.

First order derivative find how private derivative:

Here the expression is called a factorial number. The factorial of the number is equal to the product of numbers from one to, that is

The second order derivative is the first derivative of the first derivative, that is

Third-order derivative:

Fourth derivative:

Note the pattern: in the numerator there is a factorial of the number, which is equal to the order of the derivative, and in the denominator, the expression to the extent per unit is greater than the order of the derivative, that is

Answer.

Example 3.

Find the value of the third derivative function at the point.

Decision.

According to table derivatives of higher ordersWe have:

In the example under consideration, that is, we get

Note that such a result could be obtained with a consistent finding of derivatives.

At a given point, the third derivative is equal to:

Answer:

Example 4.

Find the second derivative function

Decision. To begin with, we find the first derivative:

To find the second derivative, the expression for the first derivative once again is to be indifferent:

Answer:

Example 5.

Find if

Since the specified function is a product of two functions, then to find a derivative of the fourth order, it will be advisable to apply the Formula Leibniza:

We find all the derivatives and consider the coefficients with the components.

1) Consider the coefficients at the terms:

2) We will find derivatives from the function:

3) We find derivatives from the function:

Answer:

Example 6.

The function y \u003d x 2 cos3x is given. Find a third-order derivative.

Let u \u003d cos3x, v \u003d x 2 . Then, by the Labitsa formula we find:

Derivatives in this expression have the form:

(COS3X) '\u003d - 3SIN3X,

(cos3x) '' \u003d (- 3sin3x) '\u003d - 9COS3x,

(cos3x) '' '\u003d (- 9cos3x)' \u003d 27Sin3x,

(x2) '\u003d 2x,

(x2) '' \u003d 2,

(x2) '' '\u003d 0.

Consequently, the third derivative of the specified function is equal to

1 ⋅ 27Sin3x ⋅ x2 + 3 ⋅ (-9cos3x) ⋅ 2x + 3 ⋅ (-3sin3x) ⋅ 2 + 1 ⋅ cos3x ⋅ 0

27x2Sin3x-54xcos3x-18Sin3x \u003d (27x2-18) SIN3X-54XCOS3X.

Example 7.

Find a derivativen. -o order functiony \u003d x 2 cosx.

We use the Formula Leibnitsa, believingu \u003d COSX., v \u003d X. 2 . Then

The remaining members of the row are zero, since(x2) (i) \u003d 0 at i\u003e 2.

N. derivative -o order Cosino function:

Consequently, the derivative of our function is equal

Conclusion

The school studies and use so-called formulas of abbreviated multiplication: the squares and cubes of the sum and the difference of two expressions and the decomposition formula on the multipliers of the square difference, the amounts and difference of cubes of two expressions. The generalization of these formulas is a formula called the Newton Binoma formula and the formula for decomposition of the incorporates of the amount and difference of degrees. These formulas are often used in solving various tasks: on evidence of divisibility, reduction of fractions, approximate calculations. The interesting properties of the Pascal triangle, which are closely related to Binom Newton are considered.

Information on the topic is systematized in the work, examples of the tasks for the use of Newton Binoma and the formulas of the amount and difference of degrees are given. Work can be used in the work of a mathematical circle, as well as for self-study of those who are fond of mathematics.

List of sources used

1.Vilenkin N.Ya. Combinatorics. - ed. "The science". - M., 1969

2. Nikolsky S.M., Potapov M.K., Reshetnikov N.N., Shevkin A.V. Algebra and beginning of mathematical analysis. Grade 10: studies. For general education. Organizations Basic and in-depth levels - M.: Enlightenment, 2014. - 431 p.

3. Waste for statistics, combinatorics and probability theory. 7-9 cl. / Author - Compiler V.N. Student. - ed. 2nd, sn., - Volgograd: Teacher, 2009

4.Savushkina I.A., Hugaev K.D., Tishkin S.B. Algebraic equations of higher degrees / Methodological manual for listeners of the interuniversity preparatory department. - St. Petersburg, 2001.

5. Sharygin I.F. Optional course in mathematics: solving problems. Tutorial for 10 cl. high school. - M.: Enlightenment, 1989.

6.Science and Life, Binin Newton and Triangle Pascal [Electronic resource]. - Access mode: http://www.nkj.ru/archive/articles/13598/