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The Discrete Analog of the Newton-Leibniz Formula in the Problem of Summation over Simplex Lattice Points
Evgeniy K. Leinartas Olga A. Shishkina*
Institute of Mathematics and Computer Science Siberian Federal University Svobodny, 79, Krasnoyarsk, 660041
Russia
Received 18.01.2019, received in revised form 25.02.2019, accepted 20.04.2019 Definition of the discrete primitive function is introduced in the problem of summation over simplex lattice points. The discrete analog of the Newton-Leibniz formula is found.
Keywords: summation of functions, discrete primitive function, discrete analog of the Newton-Leibniz formula.
DOI: 10.17516/1997-1397-2019-12-4-503-508.
Introduction
The problem of summation of functions of a discrete argument is one of the classical problems of the calculus of finite differences. For example, the sum of the sequence of power of natural numbers was computed by Bernoulli (1713), and his studies gave rise the development of several branches of combinatorial analysis. Euler (1733) and Maclaurin (1738) independently found that required sum is expressed in terms of derivatives and integral of a given function. It was firstly proved by Jacobi (1834) (see [1,2]).
The problem of summation of functions implies the computation of the following sum
x
S (x) = YJ 9 (t) (1)
t=0
with a variable upper limit x for a given function 9 (t). Euler proposed a method which reduces the problem to solving the difference equation
f (x +1) - f (x)= 9 (x),
where f (x) is an unknown function.
In this case, sum (1) is expressed in terms of the values of the function f (x) at points 0 and x + 1 of the segment [0, x + 1]
S (x) = f (x +1) - f (0). (2)
Function f (x) is called a discrete primitive function of the function 9 (x) and formula (2) is called discrete analog of the Newton-Leibniz formula.
* [email protected] © Siberian Federal University. All rights reserved
Many works in different branches of the mathematics are devoted to various aspects of the summation problem and its applications, among the relatively recent works note [3-5].
The summation problem in several variables can be formulated in various ways. Summation of polynomial of several variables over integer points of the rational convex polytope with variable faces was studied [6,7]. The multidimensional analogue of the Euler-Maclaurin formula was obtained.
Summation of arbitrary functions over integer points of a rational parallelotop was studied [8-10]. Euler approach based on definitions of discrete primitive function and discrete analog of the Newton-Leibniz formula was used to solve the problem.
1. Formulation of the main result
Let us introduce following notations and definitions. Vectors a1,... ,an have integer coordinates aj = (a..., a?m), aj g Z, where Z is the set of integer numbers.
Columns of matrix A contains coordinates of vectors aj:
Let cone K = {x g M" : x = Aia1 + ■ ■ ■ + Anan, Aj ^ 0} generated by vectors a1,... ,an be a pointed cone, i.e., it doesn't contain lines. In this case the system of linear equations At = x has a finite number of integer non-negative solutions for any x g K and for any function p (t) = I (t1,... ,tn) we can correctly define the following function
VA (x,*)= ^ * (t). (3)
At=x,tez™
If * (t) = 1 then VA (x; 1) is a number of representatives of the vector x by vectors a1,... ,an : x = t1a1 + ■ ■ ■ + tnan, i.e, number of vector partitions. In the case * (t) = e^t'p'> function Va (x; e(t'p>) is called vector partition function (see, for example, [7]). Function VA (x; *) is a vector partition function ith weight * (t). One can say that the problem of finding vector partition function (3) is generalization of the problem of summation of function *. Indeed, if m = 1, n = 2,
A =(1,1) and * (t1,t2) = * (t1) then Va (x; *) = £ * (h) = = £ * (t1) = S (x).
ti+t2 = x ti=0
If A = (1, 1, . . . , 1) then for the vector partition function associated with the weight function * (t1,... ,tn) we obtain VA (x; *) = £ * (t), where \\t\\ = t1 + ■ ■ ■ + tn. Let us denote the left
||t||=x
part of this equality by S (x) and formulate the following problem
Find the sum of values of function * (t1 ,...,tn) over integer points of the simplex {t G : \\t\\ = x}:
S (x)= Y * (t). (4)
Htn=x,tez5
Let us note that the case * (t) = ^ (\\t\\), where ^ (t) is a function of one variable was considered [10].
Let us introduce linear shift operators on j-th variable: Sj f (t) = f (ti.,tj + 1,..., tn), S = (Si,... Sn), Sa = Sla1 ... San. The polynomial difference operator W (S) = ^ caSa acts on
a
f (t) in the following way W (S) f (t) = J2 Caf (t + a).
a
Let us consider polynomial difference operator
W (S) = n № - Sj )
Sn
Sn-
Jn
-1
(5)
Using formula (5), for n=2 and n=3 we obtain W (Si,S2) = Si — S2 and W (Si,S2,S3) = (Si — S2) (Si — S3) (S2 — S3), respectively.
A discrete primitive function f (t) of the function 9 (t), t £ Z" is a solution of the difference equation
W (S) f (t) = 9 (t) ,t £ Z+. Let nj be a projection operator on the hyperplane tj = 0:
nj f (t) = f (ti, . . .,tj-i, 0,tj+i, ...,tn).
Let us define for n ^ 2 the Newton-Leibniz operator WNL as
(6)
WNL (S, n) = JJ (Si-Kj - Sj ni)
rn— 1 zn — 2„_
¿1 Si ni
Sn- 1 n
n- 2 Si n2
n- 1 n- 2
Sn-2 n
Sn "n
-1 1
n- 1
Note that for n=2
Wnl (S, n) = (Sin2 - S2ni)
and if f (ti,t2) is a discrete primitive function of the function 9 (ti,t2) then S (x) = £ 9 (ti,t2)= £ (f (ti + 1,t2) — f (ti,t2 + 1)) =
ti+t2 = x
ti+t2=x
= f (x + 1,0) - f (0,x +1) = Wnl (S,n) f (x,x)
In the general case problem of summation (4) can be solved with the use of the following theorem.
Theorem 1. If f (t) is a discrete primitive function of the function 9 (t) then the following discrete analog of the Newton-Leibniz formula for the vector partition function S (x) associated with 9 holds true
S (x)= J2 V (t) = Wnl (S, n) f (tiM,... ,tn) |t1 =
(7)
2. Preliminary results
To prove the theorem we need some additional results, particularly, the formula for the sum of the geometric progression
(x) =
E
qt,
t1+—+tn=x,tezr!;
n1
S
2
1
n
n
2
n
n
t =x
q
where q = (qu . . ,,qn).
Let us introduce H (z; q) =
1
(1 - qiz) ... (1 - qnz) k = 1,... ,n then after expansion in a power series we obtain
where z is complex variable. If \qkz\ < 1,
(œ \ / œ \ œ
Yé1 zM ...(Yqn ztn = £ s, (x) zx.
t1 = 0 J \t1=0 J x=0
If p < mink {|qfc 1 ^ then
sq (x) = 2n~i f H (z; q)
dz
x+1 .
(8)
\z\=p
1
Function H (z; q) x+1 has poles at points z = 0, zk = qk ,k = 1,. ..,n, and residue at an infinitely distant point is equal to 0. Then by theorem of the total sum of residues [11] we have from (8) that
n 1
(x) = rGSz=q-1 H (Z) ■
k =1
(9)
Calculating this residue, we obtain from (9) that
(x) =
qk
k=1
£-
x+n-1 qk
J n / \ / J n
1 1 - j k=1 n (qk - qj)
j=1,j=k v ' j=1 j=k
(10)
j=1,j=k
Formula (10) can be obtained in another way. For example, we can find it in [12, Sec. 3.3] (without proof). The proof can be found in [13,14] as particular case of the more general result in which arbitrary integer polytope is used instead of simplex.
If q = (q1,..., qn) and W (q) is a Vandermonde determinant then
W (q[k])
qn-2 qn-3 ... 1
qn-2 qn 3
qk-1 qk-1
qn-2 qn-3
qk+1 qk + 1
qn-2 qn-3 qn qn
n (qß- qv) •
The following equality holds true
W (q) = (-1)k-1 n (qk - qj) W (q[k]).
j=1,j=k
(11)
Let us consider operators and Sv. They are permutable and for any function f (t) the following equality holds true
(Sß - 5v) f (t) = (nvöß - nßöv) f (t)
(12)
Proof of the theorem.
1
Using equality (6) and f (t) = Stf (0), we obtain S (x) = £ r (t)= £ W (S) f (t)= £ W (S) Stf (0)= I £ ^ J W (S) f (0).
||t||=x ||t||=x ||t||=x \||t||=x J
It follows from (10) and (11) that
n SX+n-1
S (x)=£ mj-v W (S) f (0)=
n SX+n-1 n
= Y< on k (S-M)^ II (Sk - Sj) W (S[k]) f (0)
tin j=i,k=j(Sk - Sv) =1;j=k
n
= J2(-I)k-1 Sn-1W (S[k]) SXf (0).
k=1
Using (12), we find
w (S[k]) SXf (0)= n S - Sv) SXf (0) =
= n (nv Sm - n^Sv) ni... [k] ...nnf (x,x,...,x) =
= WNL (S[k],n[k]) f (x,...,x). Thus, expanding determinant WNL (S,n) by first column, we obtain
n
S (x) = J2 (-1)k-i Sn-1WNL (S[k],n[k]) f (x,...,x) = Wnl (S, n) f (x,...,x).
k=1
Formula (7) is proved. □
Example. Let us find the vector partition function with weight r (t1, t2, t3) = (3 - 2t1) 2t2 3t3 for n=3. Primitive function for r is f (t1,t2,t3) = t12t23t3. According to formula (7) of Theorem 1, sum (4) is equal to S (x) = 3X+2 - 2X+2 - x - 2.
The work was supported by RFFI (grant 18-51-41011 Uzb_t) and by RFBR (research project 18-31-00232).
References
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[3] A.V.Ustinov, A Discrete Analog of Euler's Summation Formula, Mat. Notes, 71(2002), no. 5, 851-856.
[4] A.V.Ustinov, A Discrete Analog of the Poisson Summation Formula, Mathematical Notes, 73(2003), no. 1, 97-102.
[5] S.P.Tsarev, On the rational summation problem, Programming and computer software, 31(2005), no. 2, 56-59.
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[10] O.A.Shishkina, The Euler-Maclaurin formula and differential operator of infinite order, Journal of Siberian Federal University, Mathematics & Physics, 8(2015), no. 1, 86-93.
[11] B.V.Shabat, Introduction to Complex Analysis, Part 1, Science, 1976.
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Дискретный аналог формулы Ньютона—Лейбница в задаче суммирования по целым точкам симплекса
Евгений К. Лейнартас Ольга А. Шишкина
Институт математики и фундаментальной информатики Сибирский федеральный университет Свободный, 79, Красноярск, 660041
Россия
В задаче суммирования функции по целым точкам 'рационального симплекса дано определение
дискретной первообразной и найден дискретный аналог формулы Ньютона-Лейбница для суммы.
Ключевые слова: суммирование функций, дискретная первообразная, формула Ньютона-
Лейбница.
