By Hei-Chi Chan
The purpose of those lecture notes is to supply a self-contained exposition of numerous attention-grabbing formulation chanced on by means of Srinivasa Ramanujan. principal ends up in those notes are: (1) the evaluate of the Rogers-Ramanujan persevered fraction -- a end result that confident G H Hardy that Ramanujan was once a "mathematician of the top class", and (2) what G. H. Hardy known as Ramanujan's "Most appealing Identity". This e-book covers more than a few comparable effects, comparable to a number of proofs of the well-known Rogers-Ramanujan identities and an in depth account of Ramanujan's congruences. It additionally covers quite a number recommendations in q-series.
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Extra info for An Invitation to Q-Series: From Jacobi's Triple Product Identity to Ramanujan's Most Beautiful Identity
These allow one to verify explicitly that φ2 = 1. (3) (On the Bressoud-Zeilbeger bijection; cf. 1) Let λ = (λ1 , · · · , λt ) ∈ P (n − b(j)). Suppose Case II of Eq. 8) holds. , |λ | = n − b(j + 1). (4) (On the Bressoud-Zeilbeger bijection; cf. 8) satisﬁes φ2 = 1. (5) (A variation of the Bressoud-Zeilbeger bijection; cf. 1) Deﬁne ˜ φ(λ) = (t − 3j − 2, λ1 − 1, · · · λt − 1) Case I: if t − 3j − 1 ≥ λ1 (λ2 + 1, · · · , λt + 1, 1, · · · , 1) Case II: if t − 3j − 1 < λ1 , where there are λ1 − (t − 3j) copies of 1 in Case II.
Each solid dot represents an occupied level. (b) An admissible state S1 . Compared with vac, levels 13/2, 9/2, 7/2 and 5/2 are added, while levels −1/2 and −7/2 are deleted. (c) Obtaining S1 from vac(2). An admissible state S is a state that is diﬀerent from the vacuum state by ﬁnitely many levels. 1(b). Associated with each admissible state S are: • Q(S), the charge of S, which is the number of occupied levels in S relative to the vacuum state. • H(S), the total energy of S relative to the vacuum state.
1. Exercises 000˙HC 29 (6) This is due to M. Hirschhorn (1999) and will be useful in Chapter 17. 23) n=0 n(n+1)/2 . This implies 9 (1 − q n )3 = n=1 Cn (q) . 25) (−1) (2n + 1) ≡ 0 if n ≡ 2, 7 (mod 10), ⎪ ⎪ ⎪ ⎪ −2 if n ≡ 3, 6 (mod 10), ⎪ ⎪ ⎪ ⎩−1 if n ≡ 4, 5 (mod 10). 26) J2 := S1 + S8 + S3 + S6 . 27) Show that ∞ (1 − q n )3 ≡ J1 + J2 (mod 5). 28) n=1 n n Note that ≡ n an q n bn q (mod m) means an ≡ bn (mod m). −1 Also, prove that J1 and q J2 are congruent to functions of q 5 (mod 5). 29) J2 ≡ 2qB(q 5 ) (mod 5).
An Invitation to Q-Series: From Jacobi's Triple Product Identity to Ramanujan's Most Beautiful Identity by Hei-Chi Chan