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Section 6. Mathematics
Section 6. Mathematics
DOI: http://dx.doi.org/10.20534/ESR-16-11.12-34-35
Druzhinin Victor Vladimirovich, Lebedev Ivan Mihailovic, Chernichenko Ilma Evgenievna, National research nuclear University "MEPhI" Sarov Institute of physics and technology, Department of mathematics
E-mail: Sarov, vvdr@newmail.ru
The generalization of the binomial theorem for the case arbitrary geometric progression
Abstract: The General formula of construction of sums of geometric progressions in an arbitrary natural power was obtained. The properties of such sums and their applications have been considered.
Keywords: sum of the terms of a geometric progression, binomial coefficient, the polynomial theorem.
The amount of members of a geometric progression is defined by the formula
G (a;n;l)=£ak = 1
= 1 + a + a +... + a = -
a -1
a -1
(1)
The question arises, what kind of the coefficients ck (n,t ) the exponential sum will be, if we shall construct G(n;a;l) to a positive degree t
G (a ;n;t W
a I =
a -1
a -1
= fck (n,t )
(2)
If n — 1, then the answer is known, it is the binomial theorem
G (a ;1;1) = (1 + a )
(3)
Where
k
= C = t ! / (k! (t - k) isa binomial coefficient. If t = 2 then the result is also known [1, 2]
2 n n-1
(l + a + a2 +... + an ) = £a2k
-2 £ aV2 = =
metry as in (3-5), i. e. they first increase then symmetrically dent t
crease and their sum ^ck (n,t) = (n + l) . The calculation of
k=0
specific coefficients in the exponential sum in (6) for large n and is t rather cumbersome procedure. In the paper [4] was proposed formulas for the analytical calculation of ck (n,t) . In this article, we considerably simplify the computation of such coefficients and dive their output and new versions of applications.
Considering the symmetry it is necessary to calculate under the even nt only the coefficients 0 < k < (nt ) / 2 , and under odd nt — with 0 < k <(nt -1)/2 ck (n,t)
obey the rule ck (n,t) = Cnt-k (n,t). A characteristic feature of the calculation ck (n,t ) is that the entire array of coefficients {ck} is divided into blocks of (n + l) coefficients in each bloc, and each bloc ck (n,t) should be calculated by its formula. Total necessary number of blocs is the d . If nt is even and the (nt + 2) not divis-
nt + 2
k1 (<k2 )=0 1 - k )
(4) d =
ible by (2n + 2) ,then d = nt + 2
For example,
+ 1. If the fold is, then
(l + a + a2 + a3) = 1 + 2a + 3a2 + 4a3 + 3a4 + 2a5 + a6. (5)
View of the square of the sum of a geometric progression G (n;a;2) easy to remember: first, the coefficients in power sum linearly increases through «1», peak is equal to (n + l)an, and then linearly decrease. At higher powers of t contemporary mathematics proposes to use the polynomial theorem [3; 4]
2 (n +1) than X. When nt is odd d = k
2 (n +1)
. Here [x ] gives maximum the integer not greater
nt +1
1 or d =
nt +1 2 (n +1)
G (n,t ) =
V k=0
t !
tn
= Tfk (n,t)ak =
-Is" ■ ah ■ a2s2 ...-a
(6)
■ ■■s„
where are summed up all the sets of nonnegative integers (s 0; Sp...; Sn^ are summed are summed provided S0 + s. +... + Sn = t. The coefficients c, (n,t ) have the same sym-
2 (n +1)
If - = s, then ck (n,t) is in the S block, 0 < S < d. Because
n +1 \
the coefficients of the binomial theorem, in our notation G (a; , are known, and it is included in the General scheme, the output of the calculation ck (n,t ) we shall spend on it. It turns out that binomial coefficients can be considered in such a complex scheme
In (7) p number blocks that stand in front of k and the block containing ck (l,i). Direct calculation ck (l,f), of course, be much simpler: ck (1, t) = t !/ ((t - k)kl),but it is difficult option provides the key to calculate the coefficients ck
(n ,t ) in the General
k=0
k=0
k=0
k
a
k=0
k=0
The generalization of the binomial theorem for the case arbitrary geometric progression
case. We show an example of the use of the formula (7). Have n — 1 such a sum from the selection of the blocks and their numbering
(1 + a )8 =l + 8a + 28 a2 + 56a 3 + 0 1 1 (8) +70a4 + a6a5 + 28a6 + 8a7 + a8.
2-V-' 7-V-' ^
2 3 4
Each block of coefficients consists of(n + 1) = 2 numbers. In (8) are only four blocks, but in the calculation it is necessary to take the first three of the block with p = {0;1;2}. There are nt + 2 = 10 , 2(n + l) = 4 , d — 3. Indeed, the last increasing the coefficient c 4 (1,8) = 70 is included in the block with p — 2. Thus,
, (.,8MZ(->r (4+7-2p)! -
= 8
p=0 (t - p)!(p)![4 - 2p ]!
9-10-11 8 ■ 9 7 "
— + -
2
= 70.
2 ■ 3 ■ 4 2 2 _
Generalizing equation (7) and using the method of mathematical induction, we obtain a General expression for the coefficients ck (n,t) in G(a;n;t)
(k +1 -1 - np - p )!
Xn,t )=t E(-i)p
p=o (t-p)l(p)l[k - p - up] The calculation of these ratio can be written like this
(9)
p=0
ft + k -1 - p - pn
{PA t -1
Formula (9) for ck (n,t ) checked our numerical calculations for a large number of parameters and is new in the theory of numbers. In monographs and handbooks [3-6] it is not discovered.
Lets show an application of formulas (9-10) in the following
example. We need to calculate G(a;3;3) = (l + a + a2 + a3) . Opening parenthesis in powers of ak, we get «10» terms. It is sufficient to calculate the coefficients ck (n,f ) only for the first five components, since five others repeat the first five coefficients in reverse order. Each bloc contains (n + l) = 4 terms. The calculation will involve two blocks. The coefficients of the zero bio ck (n = 3,t = 3) (9) is equal to ck (3;3) = (k + 2)!/2-k! c 0 = 1;c1 = 3;c2 = 6;c3 = 10.
The last coefficient c 4 is included in the block with s = p = 1, and it is calculated according to the formula with two members and c 4 = 6!/2 ■ 4!- 3 = 12. Thus,
G (a;3;3) = 1 + 3a + 6a2 + 10a3 + V ^ (11) +12a4 + 12a5 + 10a6 + 6 a7 + 3a8 + a9.
Lets discuss the issue of the use of the sums G (a;n;f ). Since they are a generalization of the binomial theorem in the case when
References:
(10)
there are more than two terms in the basis, then, considering the vast number of applications of the binomial theorem, the potential application of new amounts is wide. Look at some of them. 1. Calculation of improper integrals of the form
i
x"+1 -
x -1
dx =
:(n,t )
k +1
- + C.
(12)
2. There is the option of folding the exponential amount by type of the coefficients in a compact form. When we see (l + 2a + a2 )
then we immediately write (a +l) . Now our possibilities are expanding. For example,
1 + 5a + 15a2 + 30a3 + 45a4 + 51a5 + 45a6 +
+30a7 + 15a8 + 5a9 + a10 = (l + a + a2 )5
3. Differentiation of G (a ; n; t ) with respect to « a » gives a new analytically calculated amounts
dG (a;n;t )
= t
a
da
-1 ^
■ = ^kck (n,t)ak-1 =
k=1
(n + l)(a- l)an -an+
(13)
a -1 ) (a -1)
For example, differentiating G (a;3;2) we find
1 + 3a + 6a2 + 6a3 + 5a4 + 3a5 =
(a4 -l)(3a4 -4a3 +1)
(14)
(a -1)3
When a = 2 both sides of equation (14) are «255».
4. There are cases analytical solutions of algebraic equations of high degree. For example the equation of the fifth order
3a5 + 5a4 + 6a3 + 6a2 + 3a -1359 = 0,
if you look at the equality (14) has a root a = 3. Cases may arise with allocating a block G (a; n; f ) in varying degrees. For example the equation a4 + 2a3 — 3a2 — 2a —12 = 0 can be rewritten as a quadratic equation G (a; 2; 2) — 6G (a; 2; l) — 7 = 0 with roots « 7 » and « —1 ». G (a; 2; l) = 7 gives a2 + a — 6 = 0; a1 = 2; a2 =—3. The second number « — 1 » yields the complex value a.
5. The addition of increasing amounts of geometric progressions made new calculated amount. For example, we put G (a; n; l)
£G (a;n;l) = X(N - k ) =
n=0 k=0
= aN+2 +(N + 2 )a-(N +1)
= (0-7
(15)
Knowing the coefficients in G(a;n;f) we can add these polynomials and get the calculated amount.
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