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14
Section 3. Mathematics
DOI: http://dx.doi.org/10.20534/AJT-16-9.10-15-14
Druzhinin Victor Vladimirovich, Strahov Anton Viktorovich National research nuclear University «MEPHI» Sarov physical-technical Institute Department of mathematics
E-mail: vvdr@newmail.ru
Full integral formula for sums of generalized arithmetic progressions
Abstract: The formula given for the calculation of the sums with bases, forming an arithmetic progression, in the same degree with arbitrary first term. The results are for the degrees from "1" to "4". Differential and recurrence relations founded for such amounts and made the application.
Keywords: the sum of the terms of an arithmetic progression, the amount of a single-stage numbers, Bernoulli numbers.
Consider the sum of the following view
S(n;m;t;a+ (k -l)m) = a' +(a + m)' +(a + 2m)'...
k =1
+(a + (n - l)m J, (1)
which we call the full sum of the generalized arithmetic progression (SGAP) as the base of the numbers in the sum to form the AP. General properties of such sums: S(n;m;0;a) = n;S(l;m;t;a) = a'. In (1) m and a random number, t is a nonnegative integer degree. Theory of calculation of such amounts has a long history. When m = a = 1 they worked with 1617 Johann Faulhaber, who published the first book of the calculations of these amounts with powers up to "11" and after he published a second book with powers up to "17". Pierre Fermat to calculate such amounts suggested the squad-squares and cube-cubes in the letter to Mersenne. The main contribution made Jacob Bernoulli, who introduced the polynomials in his name. Later Euler found the generating function for the Bernoulli polynomials, and Appel indicated derivative for polynomials of Bernoulli. The calculation of these amounts worked well as Jacobi and other mathematicians [1]. In modern literature SGAP with m = a = 1 is calculated using the Bernoulli polynomials Bt (n) by the formula
S (n;1;t;l) = ^ [Bt+1 (n +1)-BM ]. (2)
These polynomials are found by expansion in a power series generating function ofthe Euler
efik! b (n ).
e -1 t=o t !
(3)
Bt - Bernoulli numbers, Bt = Bt (0). For small values of t < 10 these polynomials we can take their spreadsheets [2]:
B0 (x) = r;£j(x) = x -I';B2 (x) = x2 - x +1;
3 1 1
B3 (x)e x3 — x2 +-x; B. (x) = x - 2x3 + x2--;
3W 2 2 30
B5 (x ) = x5 - 5 x4
5 x 3 _ 1 x 3x 6 x
For large values of t it is necessary to paint a number (3) next. The data to calculate the value of (1) with an arbitrary interval m and an arbitrary first element a ^ 1
using the Bernoulli polynomials do not exist. There are
also other simple approaches to their analysis. In our article [3] we proposed a faster way of calculation of such amounts for arbitrary n,m ,t, but a = 1, using the integrals
S (n;m;t +1;1) = < m (t + ï))s(ç;m;t;ï)dç\+ (4)
n<! 1 - m(t + l)jS(Z;m;t;l)dZ
In this article we generalize this formula to the case of arbitrary initial number a, i. e. give the full integral formula for calculating the sums S (n ;m;t ;a)
S (n;m;t + l;a )= a'+1 p (t + l)Js fc; +
+ a'jl - m (t + l)jS m;t; l\dÇ
m a
(5)
Section 3. Mathematics
The relation (5) obtained by the method of mathematical induction and is new, as in reference books and monographs [1,2,4,5], we can't find him. When a = m = 1 (5) gives the same results as the Bernoulli polynomials (3). Show the effect of this ratio for t = 1. As S (n;l;0;l) = n, then
n n ( m1 S (n; m;l;a ) = £(a + (k - l)m ) = mfedZ + an 1--
k=1 0 \ a 0
mn2 (2a - m )n = 2 2 .
When a = m = 1, we (n;1;1;l) = n(n +1)/2 well-known formula for the sum of natural numbers. We give some specific formulas for the calculation of these amounts for the subsidiary (5)
(6)
2 3
m n
/ \ m u / \ u
S (n;m;2;a) =--+ m (2a - m)--+
3 2 ;
(6a2 - 6am + m2 )
+--- n
6
3 4 3
„ \ mn \ n
S(n;m;3;a) = ——+ m (2a-m)— +
(7)
m
+-
(6a2 - 6am + m2 ) 2 a(2a2 - 3ma + m2 )
(8)
4
2
n;
m 4n5
r \ inn 3 / \ n
S(n;m;4;a) =—-—+ m (2a - m)— +
+
m2 (6a2 - 6am + m2 )
n + m
(2a3 - 3a2
m + am2 )n2 +
+
(30a4 - 60a3m + 30a2m2 - m4 )
30
n.
(9)
For example, from (8) it is possible to obtain a new equality at m = ^ and a = 2, t = 2
J l ^ » (3 k Y 2n3 + 21n2 + 73n . N
s I n;1;2;2 J-S(3+2 J--14-. (10)
In S (n;m;t;a) the first term is always equal to (m'n'*1)/(t +1), second (m'~l (2a -m)n')/2 ,and the third (t(m^rf-)(6a2 -6am + m2)/12). If we take m = a = 1, these results coincide with the formulas of Bernoulli [2].
Next, we write the analyzed amount in the form of the polynomial
t+1
S(n;m;t;a) = ^c„ (m;t;k;a)nk. (11)
k=1
Between the coefficients cn (m;t;k;a) of (5) implies the following recurrence relation
, , N m(t + l)c (m;t;k;a) , ,
c (m;t + 1;k + 1;a) = ^-/ n V . (12)
nV ' (k +1)
The last coefficient cn (m;t + 1;1;a) is the equality
cn (m;t +1;1) = a'-^cn (m;t + 1;k). (13)
k=2
Also from (5) differential recurrent ligament
1
f
dS (n ;m;t + 1;a )
dn
= S (n;m;t;a ). (14)
m(t+1) ^
The formula S(n;m;t;a) you can search not only by integration, but also using the recurrence relation (12,13,14).
The sum S(n;m;t;a) can be differentiated by the variable m. Derivative with respect to mp -th order allows to calculate analytically the sum of view
" ,, \tt t, \ v-p (t- p)! dPS(n;m;t;a) . . g(k-l) (a + (k-1)m) {dinP '. (15)
For example, the first derivative gives
dS (n;m;t ;1)
= t±(k -1)(1 + (k - l)m)t-1. (16)
dm k=2
Indeed, when m = 2,n = 4, t = 3,a = 4
dS(4;m;3;4)_ 3m2n4
3 (32 + 2 ■ 52 + 3 ■ 72 ) =
-(4m - 3m2 ))-
dm 4
(6 - 12m + 3m2 ) 2 (2m - 3)
n = 618.
4 2
The sum S (n;m;t;a) can be differentiated on
the variable a
dS (n;m;t;a) , .
—1-- = tS(n;m;t - l;a).
da V 7
The sum S (n;m ;t ;a) can also integrate on the variable m. The result is a new ratio
m +
[a + (k - l)m +1 -1 (t + l)(k -1)
= \S (n Z;t ;a )dZ.(l7)
For example, taking m = 2,n = 3,t = 2 and proindeksirovat S (n;m;2;l) for m from zero to « m » obtained by direct calculation of the left part of the number «31,33..» and in the right part of the function
jS (n;Z;2;l)dZ =
3 3
m n
2 2 m n
m 3n2
m n m n
+ mn--+--,
2 18
which gives the same number.
The above formula for SGAP can drastically reduce the computer calculations for large amounts, analytically to work with amounts ofa new kind, with sign-alternating amounts or amounts with a predetermined order of the «+» and «—».
Using the formula (5) are solved some difficult Diophantine and algebraic equations. For example, the algebraic equation of the fifth order
768x5 - 640x3 + 112x -19680 = 0 (18) is the coefficients from unknown coincides with SGAP with t = 4,m = 4,a = 2, and one solution is just x = 2.
k=2
0
In the study of solutions to Diophantine equations The author thanks the editorial of the journal
with a large number of unknowns we found a long "Proceedings of Institute of mathematics and mechanics sequence of numbers like amounts of Farm Ur RAS", recognized the basic formula of article (4) are
23 + 43 + 73 +103 +133 +143 + B43 +173 + 203 + 233 + correct and the obtained for the first time, and also thanks
+ 233 + 243 + 283 + 353 + 463 + 533 + 613 + 703 = 983 . a member of.-Q. RAS A. A. Makhneva for reviewing the
manuscript and valuable comments.
References:
1. Prasolov V. V., Polynomials. MZIMA, -2003, -P. 131.
2. Graham, Z., Knuth, D., Patashnik O. Concrete mathematics, -Moscow, "Mir", -P. 313, - 1998.
3. Druzhinin V. V.//NTVP, -No. 5, - P. 18-20, - 2016.
4. Gradshteyn I. S., Ryzhik I. M. Tables of integrals, sums, series and products. GIMPL, - Moscow, - 1962. - P. 15-16.
5. Korn G., Korn T. Handbook of mathematics, science, GIMPL, - M., - 1974, - P. 31, 135.
