Time-Dependent Billiards

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http://ijmms.hindawi.com

A NEW PROOF OF SOME IDENTITIES OF BRESSOUD

ROBIN CHAPMAN Received 18 October 2001

We provide a new proof of the following two identities due to Bressoud:_N

m=0q^m²_N m

=

_∞

m=−∞(−1)^mq^m(5m+1)/2 ₂_N N+2m

, _N

m=0q^m²^+m_N m

=(1/(1−q^N+1))_∞

m=−∞(−1)^m× q^m(⁵^m+³^)/² _2N+2

N+2m+2

, which can be considered as ﬁnite versions of the Rogers-Ramanujan identities.

2000 Mathematics Subject Classiﬁcation: 05A19.

In [1], Bressoud proves the following theorem, from which the Rogers-Ramanujan identities follow on lettingN→ ∞.

Theorem1. For each integerN≥0, N

m=0

q^m² N

m

= ^∞

m=−∞

(−1)^mq^m(⁵^m+¹^)/² 2N

N+2m

, N

m=0

q^m²^+m N

m

= 1 1−q^N+1

∞

m=−∞(−1)^mq^m(5m+3)/2

2N+2 N+2m+2

.

(1)

Here,

N m

=







(q)N

(q)m(q)N−m if 0≤m≤N;

0 otherwise

(2)

denotes a Gaussian binomial coeﬃcient, where we adopt the standardq-series notation

(q)n= n j=1

1−q^j

. (3)

We give an alternative proof ofTheorem 1by showing that the left and right sides of (1) satisfy the same recurrence relations.

Deﬁne, for integersaandN≥0, Sa(N)=

N n=0

qⁿ²^+an N

n

. (4)

Lemma2. For each integerN≥1and eacha,

Sa(N)=Sa(N−1)+q^N+aSa+1(N−1), (5) Sa(N)=Sa+1(N−1)+q^a+1Sa+2(N−1). (6)

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Proof. Using the identity N

n

=q^N⁻ⁿ N−1

n−1

+ N−1

n

(7) gives

Sa(N)=q^N N n=1

qⁿ²^+(a−1)n N−1

n−1

+

N−1

n=0

qⁿ²^+an N−1

n

=q^N

N−1 n=0

q⁽ⁿ⁺¹⁾²^+(a−¹⁾⁽ⁿ⁺¹⁾ N−1

n

+Sa(N−1)

=q^N+aSa+1(N−1)+Sa(N−1).

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On the other hand, using the identity N

n

= N−1

n−1

+qⁿ N−1

n

(9) gives

Sa(N)= N n=1

qⁿ²^+an N−1

n−1

+

N−1 n=0

qⁿ²^+(a+1)n N−1

n

=

N−1 n=0

q⁽ⁿ⁺¹⁾²^+a(n+¹⁾ N−1

n

+Sa+1(N−1)

=q^a+1Sa+2(N−1)+Sa+1(N−1).

(10)

We now equate (5) and (6).

Lemma3. For integersN≥0and eacha, Sa(N)+

q^N+a+1−1

Sa+1(N)−q^a+1Sa+2(N)=0. (11) Proof. Equating (5) and (6) gives

Sa(N−1)+

q^N+a−1

Sa+1(N−1)−q^a+1Sa+2(N−1)=0 (12) forN≥1. ReplacingNbyN+1 gives

Sa(N)+

q^N+a+1−1

Sa+1(N)−q^a+1Sa+2(N)=0. (13)

We will use thea=0 case ofLemma 3which is S0(N)+

q^N+1−1

S1(N)−qS2(N)=0. (14) Clearly,Sa(0)=1 for alla. Also, forN >0, (5) gives

S0(N)=S0(N−1)+q^NS1(N−1) (15)

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and, together with (14), gives

S1(N)=S1(N−1)+q^N+¹S2(N−1)

=S1(N−1)+q^N

S0(N−1)+

q^N−1

S1(N−1)

=q^NS0(N−1)+

q^2N−q^N+1

S1(N−1).

(16)

Together with the initial conditionsS0(0)=S1(0)=1, (15) and (16) completely deﬁne S0(N)andS1(N)forN≥0.

We now gather some consequences of these recurrences which will be used later.

Lemma4. ForN≥2, S0(N)=

1+q²^N⁻¹

S0(N−1)+q^N 1−q^N

S1(N−2); (17) and forN≥1,

S1(N)=q^NS0(N)+ 1−q^N

S1(N−1). (18) Proof. First of all, from (15) and (16), we have

S1(N)−q^NS0(N)= 1−q^N

S1(N−1) (19)

and so, forN≥2,

S1(N−1)−q^N−1S0(N−1)=

1−q^N−1

S1(N−1). (20) Hence, by (15) again,

S0(N)=S0(N−1)+q^NS1(N−1)

=S0(N−1)+q^N

q^N−¹S0(N−1)+

1−q^N

S1(N−2)

=

1+q^2N−1

S0(N−1)+q^N 1−q^N

S1(N−2),

(21)

and also by using (16),

S1(N)=q^NS0(N−1)+

1−q^N+q^2N

S1(N−1)

=q^N

S0(N)−q^NS1(N−1) +

1−q^N+q²^N

S1(N−1)

=q^NS0(N)+ 1−q^N

S1(N−1).

(22)

The recurrences (17) and (18) with the initial conditionsS0(0)=S1(0)=1,S0(1)= 1+qdeﬁneS0(N)andS1(N)uniquely forN≥0.

Let

B0(N)=

m

(−1)^mq^m(⁵^m+¹^)/² 2N

N+2m

,

B1(N)=

m

(−1)^mq^m(⁵^m+³^)/²

2N+2 N+2m+2

(23)

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denote the sums appearing on the right sides of the identities inTheorem 1. Setting r=N+2min the deﬁnition ofB0(N)gives

B0(N)=

r≡N (4)

q^(5/8)(r^−N)²^+(1/4)(r^−N) 2N

r

−

r≡N+2(4)

q^(5/8)(r^−N)²^+(1/4)(r^−N) 2N

r

=q⁻¹^/⁴⁰

r≡N (4)

q⁽⁵^/⁸^)(r−N⁺¹^/⁵⁾² 2N

r

−

r≡N+2(4)

q⁽⁵^/⁸^)(r^−N+¹^/⁵⁾² 2N

r

. (24) This suggests the notation

A(M,k,b)=

2r≡M+k (8)

q(5/8)(r−M/2+b)²

M r

(25) so that

q^1/40B0(N)=A

2N,0,1 5

−A

2N,4,1 5

. (26)

Of course,A(M,k,b)=0 ifM+kis odd, andA(M,k,b)depends only onM,band the congruence class ofkmodulo 8. A similar computation yields

q^9/40B1(N)=A

2N+2,2,−2 5

−A

2N+2,−2,−2 5

. (27)

We aim at showing thatB0(N)and(1−q^N+1)B1(N)satisfy the same system of recurrences asS0(N)andS1(N).

Lemma5. The following holds

A(M,k,b)=A(M,−k,−b) (28)

for eachM,k, andb.

Proof. ReplacingrbyM−r in the sum forA(M,k,b)yields A(M,k,b)=

2M−2r≡M+k (8)

q(5/8)(M/2−r+b)²

M M−r

=

2r≡M−k (8)

q(5/8)(r−M/2−b)²

M r

=A(M,−k,−b).

(29)

We now wish to produce recurrences for theA(M,k,b). Lemma6. The following holds

A(M+1,k,b)=A

M,k−1,b+1 2

+q^M/2+1/10−bA

M,k+1,b+ 3 10

, A(M+1,k,b)=A

M,k+1,b−1 2

+q^M/²⁺¹^/¹⁰^+bA

M,k−1,b− 3 10

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for eachM,k, andb.

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Proof. Using the formula M+1

r

= M

r−1

+q^r M

r

(31) in the deﬁnition ofA(M+1,k,b)givesA(M+1,k,b)=S1+S2, where

S1=

2r≡M+k+1(8)

q^(5/8)(r−M/2−1/2+b)²

M r−1

=

2s≡M+k−1(8)

q⁽⁵^/⁸^)(s−M/²⁺¹^/²^+b)² M

s

=A

M,k−1,b+1 2

, S2=

2r≡M+k+1(8)

q^r+(5/8)(r−M/2−1/2+b)²

M r

.

(32)

But

r+5(r−M/2−1/2+b)²

8 =5(r−M/2+3/10+b)²

8 +M

2 + 1

10−b. (33) Hence,

A(M+1,k,b)=A

M,k−1,b+1 2

+q^M/2+1/10−bA

M,k+1,b+ 3 10

. (34) Consequently, byLemma 5also,

A(M+1,k,b)=A(M+1,−k,−b)

=A

M,−k−1,−b+1 2

+q^M/2+1/10+bA

M,−k+1,−b+ 3 10

=A

M,k+1,b−1 2

+q^M/2+1/10+bA

M,k−1,b− 3 10

.

(35)

It is convenient to note that replacingMbyM−1 in these identities gives A(M,k,b)=A

M−1,k−1,b+1 2

+q^M/²⁻²^/⁵^−bA

M−1,k+1,b+ 3 10

=A

M−1,k+1,b−1 2

+q^M/2−2/5+bA

M−1,k−1,b− 3 10

.

(36)

Lemma7. The sumsB0(N)andB1(N)obey the recurrences B0(N)=

1+q^2N−1

B0(N−1)+q^NB1(N−2) (37) forN≥2and

B1(N)=

1−q^N+1

B1(N−1)+q^N

1−q^N+1

B0(N) (38)

forN≥1.

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Proof. We compute

A

2N,k,1 5

=A

2N−1,k+1,− 3 10

+q^N−1/5A

2N−1,k−1,− 1 10

=A

2N−2,k,1 5

+q^N−³^/⁵A(2N−2,k+2,0) +q^N⁻¹^/⁵A

2N−2,k−2,2 5

+q²^N−¹A

2N−2,k,1 5

=

1+q²^N−¹ A

2N−2,k,1 5

+q^N−³^/⁵A(2N−2,k+2,0) +q^N^−1/5A

2N−2,k−2,2 5

.

(39)

In particular,

A

2N,0,1 5

=

1+q^2N−1 A

2N−2,0,1 5 +q^N^−3/5A(2N−2,2,0)+q^N−1/5A

2N−2,−2,2 5

, A

2N,4,1 5

=

1+q^2N−1 A

2N−2,4,1 5 +q^N⁻³^/⁵A(2N−2,6,0)+q^N−¹^/⁵A

2N−2,2,2 5 +q^N⁻³^/⁵A(2N−2,−2,0)+q^N−¹^/⁵A

2N−2,2,2 5

.

(40)

Noting that

A(2N−2,2,0)=A(2N−2,−2,0), A

2N−2,2,2 5

=A

2N−2,−2,−2 5

, (41)

subtracting gives

q^1/40B0(N)=A

2N,0,1 5

−A

2N,4,1 5

=

1+q^2N−1 A

2N−2,0,1 5

−A

2N−2,4,1 5

+q^N⁻¹^/⁵

A

2N−2,2,−2 5

−A

2N−2,−2,−2 5

=

1+q^2N−1

q^1/40B0(N−1)+q^N^−1/5q^9/40B1(N−2)

(42)

and so

B0(N)=

1+q^2N⁻¹

B0(N−1)+q^NB1(N−2). (43)

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Also, A

2N+2,k,−2 5

=A

2N+1,k−1, 1 10

+q^N+1A

2N+1,k+1,− 1 10

=A

2N,k,−2 5

+q^N+1/5A

2N,k−2,−1 5 +q^N+¹A

2N,k,2 5

+q²^N+⁶^/⁵A

2N,k+2,1 5

=A

2N,k,−2 5

+q^N+¹A

2N,−k,−2 5 +q^N+¹^/⁵A

2N,2−k,1 5

+q²^N+⁶^/⁵A

2N,k+2,1 5

. (44)

Consequently,

q^9/40B1(N)=A

2N+2,2,−2 5

−A

2N+2,−2,−2 5

=A

2N,2,−2 5

+q^N+1A

2N,−2−2 5

−A

2N,−2,−2 5

−q^N+¹A

2N,2,−2 5 +q^N+¹^/⁵

A

2N,0,1

5

−A

2N,4,1 5

+q^2N^+6/5

A

2N,4,1

5

−A

2N,0,1 5

=

1−q^N+¹

q⁹^/⁴⁰B1(N−1)+q^N+¹^/⁵q¹^/⁴⁰B0(N)

(45)

and so

B1(N)=

1−q^N+1

B1(N−1)+q^N

1−q^N+1

B0(N). (46)

ByLemma 4,S0(N)and(1−q^N+1)S1(N)satisfy the same recurrences asB0(N)and B1(N). Also,S0(0)=1=B0(0),S0(1)=1+q=B0(1), and(1−q)S1(0)=1−q=B1(0).

Consequently, we deduceTheorem 1:S0(N)=B0(N)and(1−q^N+¹)S1(N)=B1(N). References

[1] D. M. Bressoud,Some identities for terminatingq-series, Math. Proc. Cambridge Philos. Soc.

89(1981), no. 2, 211–223.

Robin Chapman: School of Mathematical Sciences, University of Exeter, Exeter, EX4 4QE, UK

E-mail address:[email protected]

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Special Issue on

Time-Dependent Billiards

Call for Papers

This subject has been extensively studied in the past years for one-, two-, and three-dimensional space. Additionally, such dynamical systems can exhibit a very important and still unexplained phenomenon, called as the Fermi acceleration phenomenon. Basically, the phenomenon of Fermi acceleration (FA) is a process in which a classical particle can acquire unbounded energy from collisions with a heavy moving wall.

This phenomenon was originally proposed by Enrico Fermi in 1949 as a possible explanation of the origin of the large energies of the cosmic particles. His original model was then modified and considered under diﬀerent approaches and using many versions. Moreover, applications of FA have been of a large broad interest in many diﬀerent fields of science including plasma physics, astrophysics, atomic physics, optics, and time-dependent billiard problems and they are useful for controlling chaos in Engineering and dynamical systems exhibiting chaos (both conservative and dissipative chaos).

We intend to publish in this special issue papers reporting research on time-dependent billiards. The topic includes both conservative and dissipative dynamics. Papers dis- cussing dynamical properties, statistical and mathematical results, stability investigation of the phase space structure, the phenomenon of Fermi acceleration, conditions for having suppression of Fermi acceleration, and computational and numerical methods for exploring these structures and applications are welcome.

To be acceptable for publication in the special issue of Mathematical Problems in Engineering, papers must make significant, original, and correct contributions to one or more of the topics above mentioned. Mathematical papers regarding the topics above are also welcome.

Authors should follow the Mathematical Problems in Engineering manuscript format described at http://www .hindawi.com/journals/mpe/. Prospective authors should submit an electronic copy of their complete manuscript through the journal Manuscript Tracking System athttp://

mts.hindawi.com/according to the following timetable:

Manuscript Due March 1, 2009 First Round of Reviews June 1, 2009 Publication Date September 1, 2009

Guest Editors

Edson Denis Leonel,Department of Statistics, Applied Mathematics and Computing, Institute of Geosciences and Exact Sciences, State University of São Paulo at Rio Claro, Avenida 24A, 1515 Bela Vista, 13506-700 Rio Claro, SP, Brazil; [email protected]

Alexander Loskutov,Physics Faculty, Moscow State University, Vorob’evy Gory, Moscow 119992, Russia;

[email protected]

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