# A Variation of Rogers-Ramanujan Identities

The standard Rogers-Ramanujan identities are given by $$G(q) = \sum_{n = 0}^{\infty}\frac{q^{n^{2}}}{(q;q)_{n}} = \frac{1}{(q;q^{5})_{\infty}(q^{4};q^{5})_{\infty}}$$ and $$H(q) = \sum_{n = 0}^{\infty}\frac{q^{n^{2} + n}}{(q;q)_{n}} = \frac{1}{(q^{2};q^{5})_{\infty}(q^{3};q^{5})_{\infty}}$$ and it turns out that these are non-obvious and very difficult to prove. I got one of the proofs given by Ramanujan which I presented in my blog.

Studying more on Ramanujan's work I found that he had derived further identities of similar type given by $$G_{1}(q) = \sum_{n = 0}^{\infty}\frac{q^{n^{2}}}{(q^{4};q^{4})_{n}} = \frac{(q^{2};q^{4})_{\infty}}{(q;q^{5})_{\infty}(q^{4};q^{5})_{\infty}}$$ and $$H_{1}(q) = \sum_{n = 0}^{\infty}\frac{q^{n^{2} + 2n}}{(q^{4};q^{4})_{n}} = \frac{(q^{2};q^{4})_{\infty}}{(q^{2};q^{5})_{\infty}(q^{3};q^{5})_{\infty}}$$ These identities identities look so similar to the standard Rogers-Ramanujan and I thought that they could be obtained by simple algebraic manipulation of the standard versions. However I have been unable to establish these identities so far. Please provide any helpful suggestions or references which may help me obtain a proof of the above identities concerning $G_{1}(q)$ and $H_{1}(q)$.

One standard proof of Rogers-Ramanujan's identities relies on Bailey pairs. A similar technique can then be used to prove other so-called "Identities of the Rogers-Ramanujan's type". A survey of these techniques and identities is available here.

## Bailey Pair

Let's first recall the definition of a Bailey pair and a simplified corollary of Bailey's Lemma

Definition : $(\alpha_{n}, \beta_{n})_{n=0}^{\infty}$ is a Bailey pair (relative to 1) if $$\beta_{n} = \sum_{j=0}^{n}\frac{\alpha_{j}}{(q;q)_{n-j}(q;q)_{n+j}}$$

Lemma : If $(\alpha_{n}, \beta_{n})_{n=0}^{\infty}$, then for all $\rho_{1}$ $$\sum_{n=0}^{\infty}\frac{(-1)^{n}(\rho_{1};q)_{n}}{\rho_{1}^{n}}q^{n(n+1)/2}\beta_{n} = \frac{(q/\rho_{1};q)_{\infty}}{(q;q)_{\infty}}\sum_{n=0}^{\infty}\frac{(-1)^{n}(\rho_{1};q)_{n}}{\rho_{1}^{n}(q/\rho_{1};q)_{n}}q^{n(n+1)/2}\alpha_{n}$$

For a detailed proof of Bailey's Lemma and properties of Bailey pairs, see Bailey's article Identities of the Rogers-Ramanujan Type.

## Proof of Rogers-Ramanujan's identity

To show the similarity between these proofs, let's first prove one Rogers-Ramanujan's identity using Bailey pairs. To keep it simple, let's take $$\sum_{n=0}^{\infty}\frac{q^{n^{2}}}{(q;q)_{n}} = \frac{1}{(q;q^{5})_{\infty}(q^{4};q^{5})_{\infty}}$$ From Bailey's lemma, with $\rho_{1}\rightarrow\infty$, one obtains (PBL) : $$\sum_{n=0}^{\infty}q^{n^{2}}\beta_{n} = \frac{1}{(q;q)_{\infty}}\sum_{n=0}^{\infty}q^{n^{2}}\alpha_{n}$$ One can show that $$\beta_{n} = \frac{1}{(q;q)_{n}} \qquad \qquad \alpha_{n} = (-1)^{n}(1+q^{n})q^{n(3n-1)/2}$$ is a Bailey pair. If one plugs this Bailey pair into (PBL), one gets \begin{align} \sum_{n=0}^{\infty} \frac{q^{n^{2}}}{(q;q)_{n}} & = \frac{1}{(q;q)_{\infty}}\left[1 + \sum_{n=1}^{\infty} (-1)^{n}(1+q^{n})q^{n(5n-1)/2}\right]\\ & = \frac{1}{(q;q)_{\infty}}\sum_{n=-\infty}^{\infty}(-1)^{n}q^{n(5n-1)/2}\\ & = \frac{1}{(q;q)_{\infty}}(q^{2};q^{5})_{\infty}(q^{3};q^{5})_{\infty}(q^{5};q^{5})_{\infty}\\ & = \frac{1}{(q;q^{5})_{\infty}(q^{4};q^{5})_{\infty}} \end{align} using Jacobi Triple Product between steps 2 and 3.

## Proof of $H_{1}(q)$ identity

From Bailey's lemma, with $q \leftarrow q^{2}$ and $\rho_{1} = -\sqrt{q}$, one obtains (TBL) : $$\sum_{n=0}^{\infty}q^{n^{2}}(-q;q^{2})_{n}\beta_{n}(q^{2}) = \frac{(-q;q^{2})_{\infty}}{(q^{2};q^{2})_{\infty}}\sum_{n=0}^{\infty}q^{n^{2}}\alpha_{n}(q^{2})$$ where this time $\alpha_{n}$ and $\beta_{n}$ are explicit functions of $q$. Once again, one can show that

$$\beta_{n}(q) = \frac{q^{n}(q^{1/2};q)_{n}}{(q;q)_{2n}} \qquad \qquad \alpha_{n}(q) = \left\{\begin{array}{ll}1 & \textrm{if n=0}\\[0.5em] (-1)^{n}q^{3n^{2}/4}\left[q^{3n/4}+q^{-3n/4}\right] & \textrm{otherwise} \end{array}\right.$$

defines a Bailey pair ; in fact, this is equivalent to the Bailey pair $\mathbf{G(3)}$ from Slater's list. If one plugs it into (TBL) \begin{align} \textrm{LHS} & = \sum_{n=0}^{\infty}q^{n^{2}}(-q;q^{2})_{n}\frac{q^{2n}(q;q^{2})_{n}}{(q^{2};q^{2})_{2n}}\\ & = \sum_{n=0}^{\infty}q^{n^{2}+2n}\frac{(q^{2};q^{4})_{n}}{(q^{2};q^{2})_{2n}}\\ & = \sum_{n=0}^{\infty}\frac{q^{n^{2}+2n}}{(q^{4};q^{4})_{n}} \end{align} and, using Jacobi Triple Product between steps 2 and 3 \begin{align} \textrm{RHS} & = \frac{(-q;q^{2})_{\infty}}{(q^{2};q^{2})_{\infty}} \left[1 + \sum_{n=1}^{\infty}(-1)^{n}q^{5n^{2}/2}\left[q^{3n/2}+q^{-3n/2}\right]\right]\\ & = \frac{(-q;q^{2})_{\infty}}{(q^{2};q^{2})_{\infty}} \sum_{n=-\infty}^{\infty}(-1)^{n}q^{n(5n+3)/2}\\ & = \frac{(-q;q^{2})_{\infty}}{(q^{2};q^{2})_{\infty}} (q;q^{5})_{\infty}(q^{4};q^{5})_{\infty}(q^{5};q^{5})_{\infty}\\ & = \frac{(q^{2};q^{4})_{\infty}}{(q;q)_{\infty}} (q;q^{5})_{\infty}(q^{4};q^{5})_{\infty}(q^{5};q^{5})_{\infty}\\ & = \frac{(q^{2};q^{4})_{\infty}}{(q^{2};q^{5})_{\infty}(q^{3};q^{5})_{\infty}} \end{align} That shows that $$H_{1}(q) = \sum_{n=0}^{\infty}\frac{q^{n^{2}+2n}}{(q^{4};q^{4})_{n}} = \frac{(q^{2};q^{4})_{\infty}}{(q^{2};q^{5})_{\infty}(q^{3};q^{5})_{\infty}}$$

## Proof of $G_{1}(q)$ identity

This time, we are going to use the Bailey pair $$\beta_{n}(q) = \frac{(q^{1/2};q)_{n}}{(q;q)_{2n}} \qquad \qquad \alpha_{n}(q) = \left\{\begin{array}{ll}1 & \textrm{if n=0}\\[0.5em] (-1)^{n}q^{3n^{2}/4}\left[q^{n/4}+q^{-n/4}\right] & \textrm{otherwise} \end{array}\right.$$ equivalent to the Bailey pair $\mathbf{G(1)}$ from Slater's list. If one plugs it into (TBL) \begin{align} \textrm{LHS} & = \sum_{n=0}^{\infty}q^{n^{2}}(-q;q^{2})_{n}\frac{(q;q^{2})_{n}}{(q^{2};q^{2})_{2n}}\\ & = \sum_{n=0}^{\infty}\frac{q^{n^{2}}}{(q^{4};q^{4})_{n}} \end{align} \begin{align} \textrm{RHS} & = \frac{(-q;q^{2})_{\infty}}{(q^{2};q^{2})_{\infty}} \left[1 + \sum_{n=1}^{\infty}(-1)^{n}q^{5n^{2}/2}\left[q^{n/2}+q^{-n/2}\right]\right]\\ & = \frac{(q^{2};q^{4})_{\infty}}{(q;q)_{\infty}} \sum_{n=-\infty}^{\infty}(-1)^{n}q^{n(5n+1)/2}\\ & = \frac{(q^{2};q^{4})_{\infty}}{(q;q)_{\infty}} (q^{2};q^{5})_{\infty}(q^{3};q^{5})_{\infty}(q^{5};q^{5})_{\infty}\\ & = \frac{(q^{2};q^{4})_{\infty}}{(q;q^{5})_{\infty}(q^{4};q^{5})_{\infty}} \end{align} That shows that $$G_{1}(q) = \sum_{n=0}^{\infty}\frac{q^{n^{2}}}{(q^{4};q^{4})_{n}} = \frac{(q^{2};q^{4})_{\infty}}{(q;q^{5})_{\infty}(q^{4};q^{5})_{\infty}}$$

• Thank you. I find q-series fascinating and befuddling. I have been reading Berndt's "Number Theory in the Spirit of Ramanujan" and gain about $\epsilon$ understanding each time I look inside. – marty cohen Oct 6 '13 at 20:01
• Thanka a lot Tristan. I have not yet studied about Bailey pairs, but from your proof above it looks like they are quite interesting and powerful in proving various q-series identities. Also thanks for providing the references where I can study more of this stuff. – Paramanand Singh Oct 7 '13 at 3:34