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Zbl 0911.11024
Shen, Li-Chien
On an identity of Ramanujan based on the hypergeometric series $_2F_1(\frac{1}{3},\frac{2}{3};\frac{1}{2};x)$. (English)
[J] J. Number Theory 69, No.2, 125-134 (1998). ISSN 0022-314X; ISSN 1096-1658/e

Here is the author's introduction: ``Recently {\it B. C. Berndt, S. Bhargava} and {\it 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.'' \par And here is the identity of the title: Theorem. For $0\leq q<1$, let $a= a(q)= \vartheta_3(q) \vartheta_3(q^3)+ \vartheta_2(q) \vartheta_2(q^3)$, $c= c(q)= \frac 12 a(q^{1/3})- \frac 12 a(q)$ and $h= (c^3/a^3)$; and define $$az= \int_0^\varphi {}_2F_1 \bigl( \tfrac 13,\tfrac 23, \tfrac 12;h\sin^2 t \bigr)dt.$$ Then $$\varphi=z+3 \sum_{n=1}^\infty \frac{q^n\sin 2nz}{n(1+q^n+q^{2n})}.$$ The author's elegant proof employs the following three identities: $$\frac{d\varphi}{dz}= 1+\frac{3i}{2} \Biggl\{ \frac{\vartheta_4'}{\vartheta_4} \biggl(z+ \frac{\pi\tau}{6} \biggl| \tau\biggr)- \frac{\vartheta_4'}{\vartheta_4} \biggl(z- \frac{\pi\tau}{6} \biggl| \tau\biggr)\Biggr\}= -2+\frac{3i}{2} \Biggl\{ \frac{\vartheta_1'}{\vartheta_1} \biggl(z+ \frac{\pi\tau}{3} \biggl| \tau\biggr)- \frac{\vartheta_1'}{\vartheta_1} \biggl(z- \frac{\pi\tau}{3} \biggl| \tau\biggr)\Biggr\},$$ $$\frac{d^2\varphi}{dz^2}= \frac{3i}{2} \Biggl(\wp \biggl(z- \frac{\pi\tau}{3} \biggr)- \wp\biggl(z+ \frac{\pi\tau}{3} \biggr)\Biggr), \qquad \text{ and}$$ $$\sin\varphi \cos\varphi= \frac{1}{4i} \Biggl\{ e^{4iz} \Biggl( \frac{\vartheta_1(z+\pi\tau/3)} {\vartheta_1(z-\pi\tau/3)} \Biggr)^3- e^{-4iz} \Biggl( \frac{\vartheta_1(z-\pi\tau/3)} {\vartheta_1(z+\pi\tau/3)} \Biggr)^3\Biggr\},$$ to show that the theorem is equivalent to: $$\multline e^{4iz} \Biggl( \frac{\vartheta_1(z+\pi\tau/3)} {\vartheta_1(z-\pi\tau/3)} \Biggr)^3- e^{-4iz} \Biggl( \frac{\vartheta_1(z-\pi\tau/3)} {\vartheta_1(z+\pi\tau/3)} \Biggr)^3= \frac{9}{4c^3(q^{2/3})} \Biggl(\wp \biggl(z- \frac{\pi\tau}{3}\biggr)- \wp \biggl(z+ \frac{\pi\tau}{3}\biggr)\Biggr)\\ \times \left\{ 2a(q^{2/3})- 2+\frac{3i}{2} \Biggl[ \frac{\vartheta_1'}{\vartheta_1} \biggl(z+ \frac{\pi\tau}{3}\biggr)- \frac{\vartheta_1'}{\vartheta_1} \biggl(z- \frac{\pi\tau}{3} \biggr)\Biggr] \right\}. \endmultline$$ To prove this identity, it is sufficient to show that the coefficients of the Laurent series expansions corresponding to the terms $(z-\pi\tau/3)^{-n}$, $n=1,2$ and 3, are equal on both sides, since each side is an elliptic function with the same periods and value at 0.
[M.Sheingorn (New York)]

MSC 2000:: *11F27 Theta series; Weil representation
33C05 Classical hypergeometric functions
33E05 Elliptic functions and integrals

Keywords: identity of Ramanujan; classical elliptic functions; coefficients of the Laurent series expansions

Citations: Zbl 0843.33012

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