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On an identity of Ramanujan based on the hypergeometric series 2 F 1 (1 3,2 3;1 2;x). (English) Zbl 0911.11024

Here is the author’s introduction: “Recently B. C. Berndt, S. Bhargava and F. Garvan [Trans. Am. Math. Soc. 347, 4163-4244 (1995; Zbl 0843.33012)] provided the first proof to an identity of Ramanujan. Their proof, which is based on various modular identities, is quite difficult and complicated. In this paper, we give a much simpler proof of this identity by converting it into an identity involving the classical elliptic functions and establishing the identity by comparing their Laurent series expansions at a pole.”

And here is the identity of the title: Theorem. For 0q<1, let a=a(q)=ϑ 3 (q)ϑ 3 (q 3 )+ϑ 2 (q)ϑ 2 (q 3 ), c=c(q)=1 2a(q 1/3 )-1 2a(q) and h=(c 3 /a 3 ); and define

az= 0 ϕ 2 F 1 1 3 , 2 3 , 1 2 ; h sin 2 tdt·

Then

ϕ=z+3 n=1 q n sin2nz n(1+q n +q 2n )·

The author’s elegant proof employs the following three identities:

dϕ dz=1+3i 2{ϑ 4 ' ϑ 4 (z+πτ 6τ-ϑ 4 ' ϑ 4 z - πτ 6 τ=-2+3i 2{ϑ 1 ' ϑ 1 (z+πτ 3τ-ϑ 1 ' ϑ 1 z - πτ 3 τ,
d 2 ϕ dz 2 =3i 2 z - πτ 3 - z + πτ 3,and
sinϕcosϕ=1 4ie 4iz ϑ 1 (z+πτ/3) ϑ 1 (z-πτ/3) 3 - e -4iz ϑ 1 (z-πτ/3) ϑ 1 (z+πτ/3) 3 ,

to show that the theorem is equivalent to:

e 4iz ϑ 1 (z+πτ/3) ϑ 1 (z-πτ/3) 3 -e -4iz ϑ 1 (z-πτ/3) ϑ 1 (z+πτ/3) 3 =9 4c 3 (q 2/3 ) z - πτ 3 - z + πτ 3×2a(q 2/3 )-2+3i 2ϑ 1 ' ϑ 1 z + πτ 3 - ϑ 1 ' ϑ 1 z - πτ 3·

To prove this identity, it is sufficient to show that the coefficients of the Laurent series expansions corresponding to the terms (z-πτ/3) -n , n=1,2 and 3, are equal on both sides, since each side is an elliptic function with the same periods and value at 0.

MSC:
11F27Theta series; Weil representation; theta correspondences
33C05Classical hypergeometric functions, 2 F 1
33E05Elliptic functions and integrals