Научная статья на тему 'Integral formula for the Newton’s (m, n) nominal'

Integral formula for the Newton’s (m, n) nominal Текст научной статьи по специальности «Математика»

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BINOMIAL THEOREM / THE INTEGRATION OF THE BINOMIAL

Аннотация научной статьи по математике, автор научной работы — Drushinin Victor Vladimirovich, Lebedev Ivan Mihailovich, Strahov Anton Victorovich

A new method is proposed for construction of variables degrees amounts by integrating. It is shown that in many cases it can compete with the use of the binomial theorem.

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Текст научной работы на тему «Integral formula for the Newton’s (m, n) nominal»

Section 4. Mathematics

Section 4. Mathematics

Drushinin Victor Vladimirovich, Lebedev Ivan Mihailovich, Strahov Anton Victorovich, National research nuclear University "MEPHI", Sarov physical-technical Institute E-mail: vvdr@newmail.ru

Integral formula for the Newton's (m, n) nominal

Abstract: A new method is proposed for construction of variables degrees amounts by integrating. It is shown that in many cases it can compete with the use of the binomial theorem.

Keywords: binomial theorem, the integration of the binomial.

The classical formula of the Newton's binomial is the construction of the sum of two arbitrary numbers in degree n and its has the form:

(a + b )

(1)

where Cln = n!/ (k !(n - k ) !) isa binomial coefficient. There are in (1) 2" terms, and nonrecurring — (n +1) terms. Under the Newton's (m, n) nominal we understand the amount of:

2. (a + b )3 = 3 ||(a2 + 2ab + b1 )da + ^b2db^ = a3 + 3a2b + 3b2 a + b3

3. (a + b + c)3 = 3-||(a2 + b2 + c2 + 2(ab + ac + bc))da + +j" ( + c2 + 2bc )db + ^c2db\

I =Z at+ZÏÏa

(2)

where = n. This indicates the homogeneity of functions of di-

k=1

mension n. The total amount of members right there m" integres. Squaring the sum of m numbers has a simple formula. This is the sum of the squares of the terms plus twice the product of different terms. There is only (m(m +1)/2) non-repetitive combinations. If m > 3, the modern educational and reference [1; 2] proposes to disclose the amount (2) gradually through Newton's binomial, which is very cumbersome. For example,

(a + b + c)3 = a3 + b3 + c3 + 3(a2b + ab2 + a2c + ac2 + c2b + cb2) + 6abc. (3) We propose an alternative integrated version of the calculation Newton's (m, n)nominal. First, let us formulate a general rule the taking of the integral. Let:

N(n,m1 = F(a) + F (a) + ... + Fm (am),

= a + b3 + c + 3(a 2b + ab2 + a 2c + ac2 + c2b + cb2 ) + 6abc.

Therefore, the formula (5) allows us to consistently increase the degree n of the sum m numbers, and the same term is already sorted and not repeated. The resulting coefficients are already equal to the product of binomial coefficients. For example, the number "6" in the third example is ( ■ Cj ).

The meaning of integration is to reduce basic operations. In the example 1: (a + b )2 =(a + b )(a + b )requires four operations, and the in-

(4)

V J

where F1 (a) — the sum of all summands containing a number a1, F2 (a2) — the sum of all other summands do not contain the number of a1, and so on. Thus Fm (am) one term a_ a"m. For example, in (3) F2 (b) = b3 + c3 + 3(c2b + cb2).After this split Newton's (m; n) nominal the following degrees, i. e. Newton's ( m; n +1) nominal, through the integrals written as:

tegration used only three. Any squared J requires m2, and integrating formally (m(m +1)/2) operations. In fact, such operations while integrating even less. The fact that Newton's (m, n) nominal function of the homogeneous and the number of sets of degrees tk limited. For example, when m = 3 and n = 5 possible such structures: xs; x4 y; x3 y2; x2 y2; x3 yz for cyclic permutation of the variables. Therefore, it is sufficient to take only the first integral in (5) a1 and in it to allocate separate different combinations of powers. The numerical coefficient in iront of this term go into a general formula. The following components are found by cyclic substitution.

Suppose you need to find (a + b + c )4. From the example 3 calculated:

4j(a + b + c)3 da 4b3a; 12a2bc; 6a2c2

(6)

(Zflt| =(n+Om(nm)dai+\F?.da2+\Fda■■■[+. (5)

V *-! J 1 0 0 0 J

We show the validity of formula (5) by the binomial theorem.

1. (a + b)2 = 2 J | (a + b)da + jbdbL a2 + 2ab + b2.

Making in (6) a cyclic change of variables, we get:

(a + b + c )4 = a4 + b4 + c4 + 4 (ab3 + ac3 + cb3 + bC + ba3 + ca3) +

+12 (a 2bc + b2 ac + c2ba ) + 6 (a 2b2 + a2c2 + c2b2). In the direct product would be required "81" operations, we used the "4" operation of integration. The most important thing is that the numerical multipliers to different groups of factors are immediately.

References:

1. Graham R., Knuth D., Patashnik O. Concrete mathematics. - M.: Mir, 1998.

2. Korn G., Korn T. Handbook of mathematics. - Science, GHML. - M., 1974.

k

k-1

k-1

k-1

2

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