Revisiting a Classic Identity That Implies the Rogers–Ramanujan Identities III ^†

Abstract

This is the third installment in a series of papers on a one-parameter extension of the Rogers–Ramanujan identities (this extension was discovered independently by Rogers and Ramanujan). In this paper, we report a new proof of this identity. Our key ingredient is the Bridge Lemma, an identity that connects the both sides of the one-parameter refinement, which differ significantly in terms of their complexity.

1. Introduction

The celebrated Rogers–Ramanujan identities are the following pair of identities, with $\left|q\right|<1$ (here and throughout the rest of the paper)

$$1+\sum _{n=1}^{\infty }{\displaystyle \frac{q^{n^2}}{(1-q)\cdots (1-q^n)}}=\prod _{n=1}^{\infty }{\displaystyle \frac{1}{(1-q^{5n-4})(1-q^{5n-1})}},$$

(1)

$$1+\sum _{n=1}^{\infty }{\displaystyle \frac{q^{n(n+1)}}{(1-q)\cdots (1-q^n)}}=\prod _{n=1}^{\infty }{\displaystyle \frac{1}{(1-q^{5n-3})(1-q^{5n-2})}}.$$

(2)

They are some of the most important and beautiful results in theory of partitions. For a history of these results, see [1] and Entries 38 (i) and (ii) in [2]. For a modern introduction, see [3]. Also, see the following excellent resources for their proofs: [4,5,6,7,8,9,10,11,12]. The Rogers–Ramanujan identities have transcended the boundaries of pure mathematics and found applications in diverse areas in physics, including statistical mechanics and conformal field theory (here are a few selected references [13,14,15,16,17,18,19]).

A beautiful one-parameter extension is given by

$$\sum _{k=0}^{\infty }{\displaystyle \frac{q^{k^2}}{{\left(q\right)}_k}}{a}^k=\sum _{r=0}^{\infty }{(-1)}^r{a}^{2r}{q}^{r(5r-1)/2}\left(1-a{q}^{2r}\right){\displaystyle \frac{1}{{\left(q\right)}_r{\left(a{q}^r;q\right)}_{\infty }}}.$$

(3)

Here, a is a parameter and the q-Pochhammer symbol is defined by

$${(a;q)}_n:=\prod _{k=0}^{n-1}(1-a{q}^k)\mathrm{if}n\ge 1,$$

(4)

with ${(a;q)}_0:=1$. A frequently used notation is ${\left(q\right)}_n:={(q;q)}_n$.

Equation (3) was discovered independently by Rogers and Ramanujan and their proofs appeared in the same paper [20] (which was arranged by G. Hardy). See [21] for further details. They use (3) to establish (1) and (2): this is achieved by setting $a=1$ and $a=q$, respectively, and applying the Jacobi triple product identity to the right-hand side of (3) (which converts sums to products). See [9] for details.

In our recent work [22,23], we revisited (3) in elementary manners, exploring, in particular, how various parts in the right-hand side of (3) cancel each other out to give the much simpler left-hand side. We briefly describe these results in Appendix A.

The approaches to (3) by Rogers and Ramanujan are essentially the same (a streamlined version can be found in the work of Hardy and Wright [9]). Their proofs start with the complicated right-hand side of (3), a daunting expression that seems to defy comprehension. From this, they derive a difference equation. The left-hand side of (3) emerges as a power series solution of the same difference equation.

Often, beginners who encounter (3) for the first time wonder how one can derive the right-hand side from the relatively simple left-hand side of the identity through elementary algebraic manipulations. Motivated by the significant difference in complexity between both sides of (3), it is evident that some terms need to be added to the left side. What remains very non-trivial is determining what terms should be added: there are not many clues as to how the two sides are connected or related.

In this paper, we present a new proof of (3) that addresses this question from an elementary point of view. In particular, we identify the extra terms that need to be added to the left-hand side of (3) in order to derive the right-hand side (see the Bridge Lemma in Section 2). These terms (referred to as $Z_m^{(1,2)}\left(k\right)$ below) are highly elusive because they are transformed and manipulated by non-trivial summations (e.g., see (18) and the evaluation of $J_r$ in (21)), making them unrecognizable on the right side of (3).

The details of the Bridge Lemma are provided in Section 2. A new and direct proof of (3) can be found in Section 3. We conclude with an open question in Section 4.

2. The Bridge Lemma

The main result in this section is the Bridge Lemma (Lemma 2), which provides the connection (“bridge”) between the two sides of (3).

For $k\ge 0$, define

$$Z_m\left(k\right):=Z_m^{\left(1\right)}\left(k\right)-Z_m^{\left(2\right)}\left(k\right),$$

(5)

where

• $Z_m^{\left(1\right)}\left(k\right):=0$ if $m<k$, and

  $$\begin{array}{ccc}\hfill Z_m^{\left(1\right)}\left(k\right)& :={\displaystyle \frac{q^{k^2+2kl}}{{\left(q\right)}_k}}{\displaystyle \frac{{\left(q^l;q\right)}_l}{{\left(q\right)}_l}}\hfill & \hfill \mathrm{if}m\ge k,\mathrm{with}l:=m-k.\end{array}$$

  (6)
• $Z_m^{\left(2\right)}\left(k\right):=0$ if $m\le k$, and

  $$\begin{array}{ccc}\hfill Z_m^{\left(2\right)}\left(k\right)& :={\displaystyle \frac{q^{k^2+2kl}}{{\left(q\right)}_k}}{\displaystyle \frac{{\left(q^{l+1};q\right)}_{l-1}}{{\left(q\right)}_{l-1}}}\hfill & \hfill \mathrm{if}m\ge k+1,\mathrm{with}l:=m-k.\end{array}$$

  (7)

With this understood, we have

Lemma 1.

For $k\ge 0$, the following is true:

$$Z_m\left(k\right)=\left\{\begin{array}{cc}{\displaystyle \frac{q^{k^2}}{{\left(q\right)}_k}}\hfill & ifm=k,\hfill \\ 0\hfill & \mathit{otherwise}.\hfill \end{array}\right.$$

(8)

Remark 1.

When $m=k$, $Z_m\left(k\right)$ gives the coefficient of $a^k$ on the left-hand side of (3).

Proof. 

For $m<k$, (8) follows from the definition of $Z_m^{(1,2)}\left(k\right)$ (as both of them are zero). For $m=k$, only $Z_k^{\left(1\right)}\left(k\right)$ contributes (and gives ${\displaystyle \frac{q^{k^2}}{{\left(q\right)}_k}}$). For $m\ge k+1$, (8) follows from the fact that $Z_m^{(1,2)}\left(k\right)$ are equal because of

$${\displaystyle \frac{{\left(q^l;q\right)}_l}{{\left(q\right)}_l}}={\displaystyle \frac{{\left(q^{l+1};q\right)}_{l-1}}{{\left(q\right)}_{l-1}}}.$$

□

It follows from Lemma 1 that

Lemma 2

(The Bridge Lemma).

$$a^k{\displaystyle \frac{q^{k^2}}{{\left(q\right)}_k}}=\sum _{m\ge k}{a}^m{Z}_m\left(k\right).$$

(9)

This is because only $Z_k\left(k\right)$ contributes. Note that the term on the left side of (9) is the k-th term on the left-hand side of (3).

This Lemma is very crucial in our derivation of (3), as it serves as the bridge between the two sides, which are so different in terms of complexity. Indeed, let us write (9) as

$$a^k{\displaystyle \frac{q^{k^2}}{{\left(q\right)}_k}}=\underset{\mathrm{This}\mathrm{gives}\mathrm{the}\mathrm{left}\mathrm{side}.}{\underbrace{a^k{Z}_k\left(k\right)}}+\sum _{m\ge k+1}{a}^m(\underset{=Z_m\left(k\right)}{\underbrace{Z_m^{\left(1\right)}\left(k\right)-Z_m^{\left(2\right)}\left(k\right)}}).$$

(10)

The terms in the sum over m are essential for deriving the right-hand of (3), as we will see in the next section.

The compact forms of $Z_m^{(1,2)}\left(k\right)$, as in (6) and (7), come in handy in the proof of Lemma 1. For our proof of (3) in the next section, we rewrite them as follows. Define

$${\lambda }_{r,k}\left(q\right):={\displaystyle \frac{1}{{\left(q\right)}_r}}{\left[\begin{array}{c}r\\ k\end{array}\right]}_q{(-1)}^{r-k}{q}^{(r-k)(r-k-1)/2+r^2},$$

(11)

where the q-binomial numbers (the Gaussian polynomials) are defined by

$${\left[\begin{array}{c}A\\ B\end{array}\right]}_q=\left\{\begin{array}{cc}{\displaystyle \frac{(1-q^A)(1-q^{A-1})\cdots (1-q^{A-B+1})}{(1-q^B)(1-q^{B-1})\cdots (1-q)}=\frac{{\left(q\right)}_A}{{\left(q\right)}_B{\left(q\right)}_{A-B}}}\hfill & \mathrm{if}0\le B\le A,\hfill \\ 0\hfill & \mathrm{otherwise}.\hfill \end{array}\right.$$

(12)

By applying the q-binomial theorem to the numerators (to ${\left(q^l;q\right)}_l$ and ${\left(q^{l+1};q\right)}_{l-1}$ respectively), we have

$$\begin{array}{cccc}\hfill Z_m^{\left(1\right)}\left(k\right)& =\sum _{r=k}^m{\lambda }_{r,k}\left(q\right){\displaystyle \frac{q^{(r+k)(m-r)}}{{\left(q\right)}_{m-r}}}\hfill & \hfill & \mathrm{if}m\ge k,\hfill \end{array}$$

(13)

$$\begin{array}{cccc}\hfill Z_m^{\left(2\right)}\left(k\right)& =\sum _{r=k}^{m-1}{\lambda }_{r,k}\left(q\right){\displaystyle \frac{q^{(r+k)(m-r-1)+2r}}{{\left(q\right)}_{m-r-1}}}\hfill & \hfill & \mathrm{if}m\ge k+1.\hfill \end{array}$$

(14)

See Appendix B for the idea of proof.

With the above understood, we can turn to the next section.

3. A New Proof of (3)

As we discussed in the Section 1, this proof starts from the left-hand side of (3), which is much simpler than the other side in form. Indeed, we have

$$\begin{array}{cccc}\hfill \sum _{k\ge 0}{a}^k{\displaystyle \frac{q^{k^2}}{{\left(q\right)}_k}}& =\sum _{k\ge 0}\sum _{m\ge k}{a}^m{Z}_m\left(k\right)\hfill & \hfill & (\because \mathrm{the}\mathrm{Bridge}\mathrm{Lemma})\hfill \\ \hfill & =\sum _{k\ge 0}\sum _{m\ge k}{a}^m\left(Z_m^{\left(1\right)}\left(k\right)-Z_m^{\left(2\right)}\left(k\right)\right)\hfill & \hfill & (\because (5\left)\right)\hfill \\ \hfill & =\sum _{k\ge 0}\sum _{m\ge k}{a}^m{Z}_m^{\left(1\right)}\left(k\right)-\sum _{k\ge 0}\sum _{m\ge k+1}{a}^m{Z}_m^{\left(2\right)}\left(k\right).\hfill \end{array}$$

(15)

To proceed, let us focus on the first sum in (15) (only on the sum over m):

$$\begin{array}{cccc}\hfill \sum _{m\ge k}{a}^m{Z}_m^{\left(1\right)}\left(k\right)& =\sum _{m=k}^{\infty }\sum _{r=k}^m{a}^m{\lambda }_{r,k}\left(q\right){\displaystyle \frac{q^{(r+k)(m-r)}}{{\left(q\right)}_{m-r}}}\hfill & \hfill & (\because (13\left)\right)\hfill \end{array}$$

$$\begin{array}{cc}\hfill & =\sum _{r=k}^{\infty }{\lambda }_{r,k}\left(q\right)\sum _{m=r}^{\infty }{a}^m{\displaystyle \frac{q^{(r+k)(m-r)}}{{\left(q\right)}_{m-r}}}\hfill \end{array}$$

(16)

$$\begin{array}{cc}\hfill & =\sum _{r=k}^{\infty }{\lambda }_{r,k}\left(q\right){\displaystyle \frac{a^r}{{\left(a{q}^{r+k};q\right)}_{\infty }}}.\hfill \end{array}$$

(17)

To obtain (17), we have applied to the sum over m in (16) a result due to Euler

$$\sum _{k=0}^{\infty }{\displaystyle \frac{z^k}{{\left(q\right)}_k}}={\displaystyle \frac{1}{{(z;q)}_{\infty }}}$$

(18)

(with $z=a{q}^{r+k}$).

The second sum in (15) can be handled similarly. The sum over m gives

$$\begin{array}{cccc}\hfill \sum _{m\ge k+1}{a}^m{Z}_m^{\left(2\right)}\left(k\right)& =\sum _{m=k+1}^{\infty }\sum _{r=k}^{m-1}{a}^m{\lambda }_{r,k}\left(q\right){\displaystyle \frac{q^{(r+k)(m-r-1)+2r}}{{\left(q\right)}_{m-r-1}}}\hfill & \hfill & (\because (14\left)\right)\hfill \end{array}$$

$$\begin{array}{cc}\hfill & =\sum _{r=k}^{\infty }{\lambda }_{r,k}\left(q\right)q^{2r}\sum _{m=r+1}^{\infty }{a}^m{\displaystyle \frac{q^{(r+k)(m-r-1)}}{{\left(q\right)}_{m-r-1}}}\hfill \end{array}$$

(19)

$$\begin{array}{cc}\hfill & =\sum _{r=k}^{\infty }{\lambda }_{r,k}\left(q\right){\displaystyle \frac{a^{r+1}{q}^{2r}}{{\left(a{q}^{r+k};q\right)}_{\infty }}}.\hfill \end{array}$$

(20)

Again, we have applied (18) to the sum over m in (19) to obtain the last line.

Using (17) and (20) in (15), we have

$$\begin{array}{cc}\hfill & \sum _{k\ge 0}{a}^k{\displaystyle \frac{q^{k^2}}{{\left(q\right)}_k}}\hfill \\ \hfill & =\sum _{k=0}^{\infty }\sum _{r=k}^{\infty }{\lambda }_{r,k}\left(q\right)a^r{\displaystyle \frac{\left(1-a{q}^{2r}\right)}{{\left(a{q}^{r+k};q\right)}_{\infty }}}\hfill \\ \hfill & =\sum _{r=0}^{\infty }{a}^r\left(1-a{q}^{2r}\right)\sum _{k=0}^r{\displaystyle \frac{{\lambda }_{r,k}\left(q\right)}{{\left(a{q}^{r+k};q\right)}_{\infty }}}\hfill \\ \hfill & =\sum _{r=0}^{\infty }{(-1)}^r{a}^r{q}^{r^2}\left(1-a{q}^{2r}\right){\displaystyle \frac{1}{{\left(q\right)}_r}}\underset{:=J_r=a^r{\displaystyle \frac{q^{r(3r-1)/2}}{{(a{q}^r;q)}_{\infty }}} ; \mathrm{see}\mathrm{Appendix}\mathrm{C}}{\underbrace{{\sum }_{k=0}^r{\left[\begin{array}{c}r\\ k\end{array}\right]}_q{(-1)}^r{\displaystyle \frac{q^{(r-k)(r-k-1)/2}}{{\left(a{q}^{r+k};q\right)}_{\infty }}}}}\hfill \\ \hfill & =\sum _{r=0}^{\infty }{(-1)}^r{a}^{2r}{q}^{r(5r-1)/2}\left(1-a{q}^{2r}\right){\displaystyle \frac{1}{{\left(q\right)}_r{\left(a{q}^r;q\right)}_{\infty }}},\hfill \end{array}$$

(21)

which is the right-hand side of (3). Note that we have used the definition of ${\lambda }_{r,k}\left(q\right)$ (i.e., (11)) to obtain (21). To obtain the last step, we need to establish the evaluation of $J_r$ in (21)—see Appendix C for details.

4. Conclusions

We conclude this paper with a challenge: Find a connection between our approach and proofs based on Bailey’s Lemma (e.g., see [24,25]). In particular, see [24] for a detailed proof of (3) through the framework of Bailey’s lemma.