Научная статья на тему 'On generating functions for lattice paths'

On generating functions for lattice paths Текст научной статьи по специальности «Математика»

CC BY
100
23
i Надоели баннеры? Вы всегда можете отключить рекламу.
Ключевые слова
GENERATING FUNCTIONS / LATTICE PATHS

Аннотация научной статьи по математике, автор научной работы — Chandragiri Sreelatha,

We obtained the functional equation for generating functions of a number of generalized lattice paths.

i Надоели баннеры? Вы всегда можете отключить рекламу.
iНе можете найти то, что вам нужно? Попробуйте сервис подбора литературы.
i Надоели баннеры? Вы всегда можете отключить рекламу.

Текст научной работы на тему «On generating functions for lattice paths»

UDK 519.2/.6 Sreelatha Chandragiri, E.K. Leinartas

Siberian Federal University

ON GENERATING FUNCTIONS FOR LATTICE PATHS

We obtained the functional equation for generating functions of a number of generalized lattice paths. Keywords: generating functions, lattice paths.

Let V = {/} be the set of all ordered sets J = {j1, j2, ••• }, 1 — A < ••• < ik — N, k = 0,1,2, ..., N,#J = k number of elements in the set J. Denote nj the projection operator along the y'-th coordinate axis in Mn, i.e. njx = (x1, ..., Xj-1 , 0, Xj+1, ..., xN), and its action on the function ^>(x): ZN ^ C is defined as follows:

nj<p(x) = <p{njx), j = 1, ..., N.

We denote C[<f] the ring of formal power series in the variable % = (H,..., ) and define the action of the operator nj for j = 1,..., N on the generating series O(0 = YjXEzn V(x)^x E C[<f]. If J E V denote nj = n^o — o njk composition of the operators njl,..., njk and = 1 is the identity operator. Shift operator Sj at y'-th variable Sj<p(x) = (x1,..., Xj-1 , Xj + 1, Xj+1, ..., xN).

Let the function ^(x) satisfies the basic recurrence relation of the combinatorial analysis:

^(x) — <p(x — e1) —----<p(x — eN) = 0.

The effective method of researching sequences in enumerative combinatorics is generating functions (see [2], [5], [8]). The main result is the identity which allows us to investigate many properties of combinatorial sequences.

Theorem. The generating series OOf) = "LxEzN *P(X)^X for the function <p(x): Z> ^ C satisfies the identity

YJ(—1)#1n,[(1 — a 0)^(0]= ^ (1 — a s-i))^(X)^x, (1)

}EV XE1+TS

where (I,0 = ^ + • + and I = (1, .,1).

Proof. Firstly, for an arbitrary series 0(<f) = TlxEzN <p(x)^x, operator n = EjeV(—1)#Jnj acts on the series OOf) as follows:

n: O(0 = ^ <p(xW ^ ^ <p(xW. (2)

xEli[ xEI+lg.

To prove (2), we represent the operator n in the form of a composition of the operator n = (1 — n1)(1 — n2) • (1 — nN) and considering the commutativity of its factors, we apply it to the series

O(0:

(1 — n1)(1 — n2) • (1 — Un)4>(0 = = (1 — U1)(1 — • (1 — ^N-1)[0(^) — HnN0] =

= (1 — U1)(1 —U2) • (1 —nN-1) ^ (p(x)i;X =

xEeN+TS

= (1 — U1)(1 —U2) • (1 —nN-2) ^ (p(x)ZX =

xEeN+eN-l+li

= ...= ^ <p(x)fx,

XEI+T^l

where I = e1 + e2+... +eN.

Secondly, if n,- = (1 — n^) ••• (l — nJ-1)(l — ftj+i) ••• (1 — ), then for any j = 1,...,N the following equality holds

n^o(0 = n^o(0 = ^ <p(x —^. (3)

To prove (3) we represent the operator n in the form of a composition of operators and apply it to e C: :

(1 — ni)(1 — n2) ••• (1 — nN)[^jO(^)] = = (1 — ni) • (1 — nj-i)(1 — nj+i) ••• (1 — UN)[(1 — ^0(0] = = (1 — ni) • (1 — Uj-i)(1 — nj+i) ••• (1 — Un— ^jOtf)] = = (1 — ni) • (1 — nj-i)(1 — nj+i) ••• (1 — u^^Otf)] = = n ^ <p(x — = ^ <p(x — .

Finally, we apply the operator n to the product (1 — </, 0)O(£):

n[(1 — </, 0)Otf)] = notf) — n[</, ftOtf)] = no(0 — a noo(0 =

= notf) — nifrOtf)-----n^Otf) =

= 2*e/+z£ <Kx)r — E*e/+z£ <P(x — ^X* —... — E*e/+z£ <P(x — X* =

= 2xe/+z^(x) — <KX — gi)—... — Kx — =

= [(1 — </, <5-i>Mx)K*.

We consider the particular case of the lattice paths with steps from an orthonormal basis A = (ei,..., } and we denote ^(x): the number of paths from the origin to the point x e .

Proposition 1. If ^(x) is the number of lattice paths from the origin to xe2> using steps from the set A = (ei,..., }, then its generating function 0(<f) equals to

o(0 = 1

Proof. We note that function ^(x) satisfies to the basic recurrence relation, which implies that the left side of the identity (1) is equal to 0.

Let's write the identity (1) for the two dimensional case:

(1 — — — (1 — 12)0(0, &) — (1 — ?i)0tfi, 0) + 0(0, 0) = 0.

Since ^(xi, 0) = ^(0, x2) = 1 for all nonnegative integers xi and x2, then we get

0(0, ^2)=1I1^, o(^i,0)=r-1^, 0(0,0) = 1

And finally 0(^i, <f2) = i

For N = 3 we have

(1 - - - fc, - (1 - - ?2, 0) - (1 - - ^Ж^ 0, ^з)

- (1 - ^2 - ?з)Ф(0, fc, <Тз) + (1 - 0, 0) + (1 - &)Ф(0, fc, 0)

+ (1 - ^з)Ф(0, 0, &) - Ф(0, 0, 0) = 0.

Considering two dimensional case, we get

Repeating this process we get the generating function for any N >1.

Consider the problem of counting lattice paths with steps (1,0) and (0,1) which starts at the origin and stay on or above the line y = x (see[1],[6,7]). Let /"(x, y) denote the number of paths going from (0,0) to (x, y). Obviously, the number of paths /(x, y) satisfies the difference equation

/(x, y) — /(x — 1, y) — /(x, y — 1) = —5o (x — y)/(x, y — 1), (4)

{1 j/* x = y

n' ._r ^ and x, y > 1.

0, i/ x ^ y y

We have the following initial conditions:

/(x,0) = i, x = 0,i,2,.... (5)

/(0,y) = i, y = 0,i,2,.... ( )

Let /(x, y) be the function /: Z+ ^ C and = £x,y>0/(x, y)£i £2 be the generating

function of /(x, y).

Let Fii(t) denote the diagonal power series of F(^i,^2):

iii(0 = k)tfc. fc = 0

Proposition 2. let be the generating function of the solution of (4). Then the series F(^2) satisfies the following functional equation

(1 — fc — — (1 — WFtfi, 0) — (1 — ^(0, fc) + m 0) = ^(^2).

If the solution of /(x, y) satisfies the initial conditions (5), then

, , 1 — V1 — 4t2 Fn(t> =--■

Consider the problem of counting lattice paths with steps (1,0), (0,1), (1,1) which starts at the origin and stay on or above the line y = x. Let /(x, y) denote the number of paths going from (0, 0) to (x, y). Obviously, the number of paths /(x, y) satisfies the difference equation

/0, y) — /0 — 1, y) — /0, y —1) — /0 — 1, y —1) = —^0 (x — y)/(x, y — 1). (6)

We have the following initial conditions:

/(x,0) = i, x = 0,i,2,.... (7)

/(0,y) = i, y = 0,i,2,.... ( )

Proposition 3. Let F(^i, <f2) be the generating function of the solution of (6). Then the series F(^2) satisfies the following functional equation

(1 — — k — <^2)^(^2) — (1 — ?i)Ftfi, 0) — (1 — ^2)^(0, + ^(0, 0) = fiFntfifc). If the solution of /(x, y) satisfies the initial conditions (7), then

1 — t2 — V1 — 6t2 + t4 F"(t) =-^-.

As a special case, this example includes some well-known lattice-path enumeration problems see

([1], [3], [4]).

References:

1. M. Bousquet-Melou, M. Petkovsek, Linear recurrences with constant coefficients: the multivariate case, Discrete Mathematics 225 (2000). PP. 51-75.

2. R. Stanley, Enumerative combinatorics, Volume 1, 1990.

3. P. Duchon, On the enumeration and generation of generalized Dyck words, Discrete Math. 225 (2000). PP. 121-135.

4. E.K. Leinartas, A.P. Lyapin, On the rationality of multidimensional recursive series, Journal of Siberian Federal University Mathematics and Physics. 2 (4), (2009). PP. 449-455.

5. J. Labelle, Y.N. Yeh, Generalized Dyck paths, Discrete Math. 82 (1990). PP. 1-6.

6. A. Lyapin, S. Chandragiri, Generating functions for the number of paths on multidimensional integer lattice, ICDEA (2017). PP. 82.

7. S. Chandragiri, A. Lyapin, On an identity of Chaundy and Bullard for vector partition functions, ICDEA(2018). PP. 52.

8. A.A. Kytmanov, A.P. Lyapin, T.M. Sadykov, Evaluating the rational generating function for the solution of the Cauchy problem for a two-dimensional difference equation with constant coefficients, Programming and computer software Vol. 43, No. 2 (2017). PP. 105-111.

Сведения об авторах

Sreelatha Chandragiri

PhD student

Siberian Federal University Krasnoyarsk, Russia Email: srilathasami66@gmail. com E.K Leinartas Scientific adviser, professor Siberian Federal University Krasnoyarsk, Russia Email: lein@mail.ru

Information about authors

Срелатаха Чандрагири

Аспирант

Сибирский федеральный университет Красноярск, Россия Эл. почта: srilathasami66@gmail.com Е. К. Лейнартас

Научный руководитель, профессор Сибирский федеральный университет Красноярск, Россия Эл. почта: lein@mail.ru

UDK 517.956.8 Yu.Yu. Klevtsova

Siberian Regional Hydrometeorological Research Institute Siberian State University of Telecommunications and Information Sciences

ON THE RATE OF CONVERGENCE AS t ^ +<x> OF THE DISTRIBUTIONS OF SOLUTIONS TO THE STATIONARY MEASURE FOR THE STOCHASTIC SYSTEM OF THE QUASI-SOLENOIDAL LORENZ MODEL FOR A BAROCLINIC ATMOSPHERE

It was obtained the sufficient conditions on the right-hand side and the parameters of Lorenz model for a baroclinic atmosphere with white noise perturbation for existence of a unique stationary measure of Markov semigroup defined by solutions of the Cauchy problem ^ for this system and^ for the exponential convergence of the distributions of solutions to the stationary measure as t ^ +<x>. Keywords: Lorenz model, white noise perturbation, stationary measure, rate of convergence

We consider the system of equations for the quasi-solenoidal Lorenz model for a baroclinic atmosphere

iНе можете найти то, что вам нужно? Попробуйте сервис подбора литературы.

d

— Aiu + vA2u + A3u + B(u) = g , t > 0, (1)

dt

on the two-dimensional unit sphere S centered at the origin of the spherical polar coordinates (X, 9), n n

Xe [0,2п), фе

2 2

ц = sin ф. Here v > 0 is the kinematic viscosity,

u(t, x, ra) = (ui (t, x, ra), u 2(t, x, ra)) is an unknown vector function and

g(t,x,ra) = (g1(t,x,ra),g2(t,x,ra))T is a given vector function, x = (X,raeQ, (Q,P,F) is a complete probability space,

i Надоели баннеры? Вы всегда можете отключить рекламу.