[Calculation 16] Ramanujan ‘s 1ψ1 (1-psi-1) Summation Formula

Edited by Leun Kim

In this post, we will introduce one of the famous formulas discovered by Ramanujan, which is called Ramanujan’s ${}_1\psi_1$ Summation Formula. It was first introduced by Hardy, and he called it as “a remarkable formula with many parameters”. The first published proof was given by W. Hahn [1] in 1949.

 Theorem. (Ramanujan’s ${}_1\psi_1$ Summation Formula) If $|\beta q|< |z|<1/|\alpha q|$ then \begin{align*} &1+\sum_{k=1}^\infty \frac{(\frac{1}{\alpha}; q^2)_k (-\alpha q)^k}{(\beta q^2;q^2)_k} z^k + \sum_{k=1}^\infty \frac{(\frac{1}{\beta}; q^2)_k (-\beta q)^k}{(\alpha q^2;q^2)_k} z^{-k}\\ &= \left( \frac{(-qz;q^2)_\infty (-q/z;q^2)_\infty}{(-\alpha qz; q^2)_\infty (-\beta q/z; q^2)_\infty} \right) \left( \frac{(q^2;q^2)_\infty (\alpha \beta q^2;q^2)_\infty}{(\alpha q^2; q^2)_\infty (\beta q^2; q^2)_\infty} \right). \end{align*}

Proof. Define $g$ as
$$g(z) = \frac{(-qz;q^2)_\infty (-q/z;q^2)_\infty}{(-\alpha qz; q^2)_\infty (-\beta q/z; q^2)_\infty}.$$
Since $g(z)$ is analytic in the annulus, $|\beta q|< |z|<1/|\alpha q|$, we can write $$g(z) = \sum_{k=-\infty}^\infty c_k z^k, \qquad |\beta q|<|z|<1/|\alpha q|.$$ From the definition of $g$, we can easily check that $$\tag{1} (\beta + qz) g(q^2 z) = (1+\alpha q z)g(z)$$ provided that $|\beta q|<|q^2 z|$. Comparing the coefficients of $z^k$ in the equation (1), we find that $$\tag{2} \beta q^{2k} c_k + q^{2k-1} c_{k-1} = c_k + \alpha q c_{k-1}$$ for $-\infty < k < \infty$. Evaluating $c_k$ and $c_{-k}$ from (2), we obtain $$c_k = - \frac{\alpha q (1-q^{2k-2}/\alpha)}{1-\beta q^{2k}} c_{k-1}, \qquad 1\le k < \infty$$ and $$c_{-k} = - \frac{\beta q (1-q^{2k-2}/\beta)}{1-\alpha q^{2k}} c_{-k+1}, \qquad 1\le k < \infty,$$ where we replaced $k$ by $1-k$ in (2) to get the last equality. Iterating these relations, we deduce that $$\tag{3} c_k = \frac{(-\alpha q)^k (1/\alpha;q^2)_k}{(\beta q^2; q^2)_k} c_0, \qquad 1\le k < \infty$$ and $$\tag{4} c_{-k} = \frac{(-\beta q)^k (1/\beta;q^2)_k}{(\alpha q^2; q^2)_k} c_0, \qquad 1\le k < \infty$$ respectively. Therefore we obtain $$\tag{5} g(z) = \sum_{k=-\infty}^\infty c_k z^k = c_0 \left( 1 + \sum_{k=1}^\infty \frac{ (1/\alpha;q^2)_k (-\alpha q)^k}{(\beta q^2; q^2)_k} z^k + \sum_{k=1}^\infty \frac{ (1/\beta;q^2)_k (-\beta q)^k}{(\alpha q^2; q^2)_k} z^{-k}\right).$$ Note that $g(z)$ has a simple pole at $z = - 1/\alpha q$. Multiply both sides of (5) by $1+\alpha q z$, and $z \to -1 / \alpha q$ to get $$\lim_{z \to - \frac{1}{\alpha q}} (1+\alpha q z) g(z) = \frac{(1/\alpha; q^2)_\infty }{(\beta q^2 ; q^2)_\infty} c_0,$$ where we applied Abel's Continuity Theorem (see for example [3]). By the definition of $g(z)$, we obtain $$\frac{(1/\alpha; q^2)_\infty (\alpha q^2; q^2)_\infty}{(q^2; q^2)_\infty (\alpha \beta q^2; q^2)_\infty} = \frac{(1/\alpha; q^2)_\infty }{(\beta q^2 ; q^2)_\infty} c_0$$ so that $$\tag{6} c_0 = \frac{(\beta q^2 ; q^2)_\infty (\alpha q^2; q^2)_\infty}{(q^2; q^2)_\infty (\alpha \beta q^2; q^2)_\infty}.$$ Replacing $c_0$ in (6) into (5), we obtain the theorem for $|\beta / q| < |z| < 1 / |\alpha q|$. Analytic continuation yields the theorem for $|\beta q|< |z|<1/|\alpha q|$.$\square$

References.
[1] W. Hahn, Beiträge zur Theorie der Heineschen Reihen, Math. Nachr. 2 (1949), 340-379.
[2] Bruce C. Berndt, Ramanujan’s Notebooks, Part III, Springer-Verlag, pp. 31-33.
[3] G. E. Andrews, R. Askey and R. Roy, Special Functions, Cambridge University Press, pp. 502-505.

Leun Kim

Ph.D Candidate at The University of Tokyo
I was born and raised in Daegu, S. Korea. I majored in electronics and math in Seoul from 2007 to 2012. I've had a great interest in math since freshman year, and I studied PDE in Osaka, Japan from 2012-2014. I worked at a science museum and HUFS from 2014 in Seoul. Now I'm studying PDE in Tokyo, Japan. I also developed an interest in music, as I met a great piano teacher Oh in 2001, and joined an indie metal band in 2008. In my spare time, I enjoy various things, such as listening music, blogging, traveling, taking photos, and playing Go and Holdem. Please do not hesitate to contact me with comments, email, guestbook, and social medias.

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