By D. Mumford

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45, Entry 29(i)], we ﬁnd that ϕ(−q) + ϕ(−q 5 ) = 2 (q; q)∞ f (q 3 , q 7 )f (q 4 , q 6 ) . 2). 7). 2 with a misprint corrected. 2 (p. 56). If α = 1 −2 5 and β = 1 +2 5 , then 30 1 Rogers–Ramanujan Continued Fraction – Modular Properties ϕ(q) + (i) √ (1 + 5ϕ(q 5 ) = ϕ(q) − (ii) 5)f (−q 2 ) 1 − βq n + q 2n 1 + αq n + q 2n n odd √ √ (1 − 5ϕ(q 5 ) = √ n even 5)f (−q 2 ) 1 − αq n + q 2n n even f (−q ) 1 − βq n + q 2n 1 + αq + q 2n n n odd √ 1 + βq n + q 2n , n odd f (−q 2 ) (iv) ψ(q 2 ) − q 5ψ(q 10 ) = , n odd 2 √ (iii) ψ(q 2 ) + q 5ψ(q 10 ) = , 1 − αq n + q 2n 1 + βq n + q 2n n odd .

15) have the forms, ⎧ 4u1 = −p + 2a + p2 + 4a2 − 16k − 4pa + 16b, ⎪ ⎪ ⎪ ⎨4u = −p + 2a − p2 + 4a2 − 16k − 4pa + 16b, 2 ⎪ = −p − 2a + p2 + 4a2 − 16k + 4pa − 16b, 4u 3 ⎪ ⎪ ⎩ 4u1 = −p − 2a − p2 + 4a2 − 16k + 4pa − 16b. 15) by using the same steps as explained above. 16), p= 1 , v3 q= 3 , v r = −v, and s = − 1 . , 1 3 2 1 k + − 2+ 2v 4 v 3 3 = k3 − k2 + 2 k + 2v 4v k3 − 12 1 4 1 − 3 + 8 − v2 k+ 2 v 8 v v 12 1 1 − 3 + 8 − v 2 = 0. 26) with c2 cd 2c3 + and h = d − . 24), g=f− c=− 3 , 2v d= 3 , 4v 2 and 1 8 f= − 12 1 + 8 − v2 .

Note that t is then replaced by ζ j t. Multiplying these ﬁve identities together, we deduce that 4 1 ζjt j=0 4 = j=0 −α ζjt 1 (ζ j q 1/5 )1/2 ∞ f (−q) 1 f (−q 5 ) n=1 1 + α(ζ j q 1/5 )n + (ζ j q 1/5 )2n which can be easily reduced to , 24 1 Rogers–Ramanujan Continued Fraction – Modular Properties 1 √ t 5 √ − α t 5 = 4 ∞ f 5 (−q) 1 . 2. 9) becomes 1 √ t 5 √ − α t 5 1 = q 1/2 f (−q) f (−q 5 ) ∞ k=1 1 . 1. 11). 4) in a similar manner. 11). 1). Following Ramanujan, set u = R(q), u = −R(−q), v = R(q 2 ), and w = R(q 4 ).

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