On a septuple product identity

Robin Chapman School of Mathematical Sciences University of Exeter EX4 4QE UK

               rjc@maths.exeter.ac.uk

For polynomials f and g in the variable n with integer coefficients let us define

Ω(f) = ∑_{n=−∞}^{∞}(−1)^{n}q^{f(n)}

and

Ω(f, g) = ∑_{n=−∞}^{∞}(−1)^{n}q^{f(n)}x^{g(n)}.

In [1] Farkas and Kra state and prove the following identity

Theorem 1

∏_{n=1}^{∞}(1 − q^{2n})^{2}(1 − xq^{2n−2})(1 − x^{−1}q^{2n})(1 − x^{2}q^{4n−2})

× (1 − x^{−2}q^{4n−2})(1 − x^{2}q^{4n−4})(1 − x^{−2}q^{4n})
=  Ω(5n^{2} + n)(Ω(5n^{2} + 3n, 5n + 3) + Ω(5n^{2} − 3n, 5n))
− Ω(5n^{2} + 3n)(Ω(5n^{2} + n, 5n + 2) + Ω(5n^{2} − n, 5n + 1)).
□

Their proof exploits identities involving theta functions. Foata and Han [2] gave a more elementary deduction of (1) from Jacobi’s triple product formula and Watson’s quintuple product formula. Here we give a still more elementary derivation just from Jacobi’s triple product formula.

Let P denote the product on the of the identity in Theorem 1. First observe that we can rewrite P as

∏_{n=1}^{∞}(1 − q^{2n})^{2}(1 − xq^{2n−2})(1 − x^{−1}q^{2n})(1 − x^{2}q^{2n−2})(1 − x^{−2}q^{2n}). (1)

We now use Jacobi’s triple product formula in the form

∏_{n=1}^{∞}(1 − q^{2n})(1 − xq^{2n−2})(1 − x^{−1}q^{2n}) = ∑_{k=−∞}^{∞}(−1)^{k}x^{k}q^{k^{2}−k}.  (2)

We obtain (2) by replacing q by q^{2} and k by −k in formula (1.1) of [2]. Replacing x by x^{2} in (2) gives

∏_{n=1}^{∞}(1 − q^{2n})(1 − x^{2}q^{2n−2})(1 − x^{−2}q^{2n}) = ∑_{k=−∞}^{∞}(−1)^{k}x^{2k}q^{k^{2}−k}.  (3)

Multiplying (2) by (3) and using (1) gives

P = ∑_{k,l=−∞}^{∞}(−1)^{k+l}x^{k+2l}q^{k^{2}−k+l^{2}−l}.

Thus

P = ∑_{r=−∞}^{∞} a_{r}(q)x^{r}

where

a_{r}(q) = ∑_{k+2l=r}(−1)^{k+l}q^{k^{2}−k+l^{2}−l} = ∑_{l=−∞}^{∞}(−1)^{r−l}q^{5l^{2}−(4r−1)l+r^{2}−r}.

We now evaluate a_{r}(q) by dividing into cases according to the residue of r modulo 5. If r = 5m then

5l^{2} − (4r − 1)l + r^{2} − r = 5(l − 2m)^{2} + (l − 2m) + 5m^{2} − 3m and so

a_{5m}(q) = (−1)^{m}q^{5m^{2}−3m} ∑_{n=−∞}^{∞}(−1)^{n}q^{5n^{2}+n} = (−1)^{m}q^{5m^{2}−3m}Ω(5n^{2} + n).

If r = 5m + 1 then

5l^{2} − (4r − 1)l + r^{2} − r = 5(2m − l)^{2} + 3(2m − l) + 5m^{2} − m and so

a_{5m+1}(q) = −(−1)^{m}q^{5m^{2}−m} ∑_{n=−∞}^{∞}(−1)^{n}q^{5n^{2}+3n} = −(−1)^{m}q^{5m^{2}−m}Ω(5n^{2} + 3n).

If r = 5m + 2 then 5l^{2} − (4r − 1)l + r^{2} − r = 5(l − 2m − 1)^{2} + 3(l − 2m − 1) + 5m^{2} + m and so

a_{5m+2}(q) = −(−1)^{m}q^{5m^{2}+m} ∑_{n=−∞}^{∞}(−1)^{n}q^{5n^{2}+3n} = −(−1)^{m}q^{5m^{2}+m}Ω(5n^{2} + 3n).

If r = 5m + 3 then

5l^{2} − (4r − 1)l + r^{2} − r = 5(2m + 1 − l)^{2} + (2m + 1 − l) + 5m^{2} + 3m and so

a_{5m+3}(q) = (−1)^{m}q^{5m^{2}+3m} ∑_{n=−∞}^{∞}(−1)^{n}q^{5n^{2}+n} = (−1)^{m}q^{5m^{2}+3m}Ω(5n^{2} + n).

If r = 5m + 4 then 5l^{2} − (4r − 1)l + r^{2} − r = 5(l − 2m − 2)^{2} + 5(l − 2m − 2) + 5m^{2} + 5m + 2 and so

a_{5m+4}(q) = (−1)^{m}q^{5m^{2}+5m+2} ∑_{n=−∞}^{∞}(−1)^{n}q^{5n^{2}+5n}.

But in this sum the terms for n = j and n = −j −1 cancel and so a_{5m+4}(q) = 0. It follows that

P = Ω(5n^{2} + n) ∑_{m=−∞}^{∞}(−1)^{m}(q^{5m^{2}−3m}x^{5m} + q^{5m^{2}+3m}x^{5m+3})

−Ω(5n^{2} + 3n) ∑_{m=−∞}^{∞}(−1)^{m}(q^{5m^{2}−m}x^{5m+1} + q^{5m^{2}+m}x^{5m+2})

= Ω(5n^{2} + n)(Ω(5n^{2} − 3n, 5n) + Ω(5n^{2} + 3n, 5n + 3))

−Ω(5n^{2} + 3n)(Ω(5n^{2} − n, 5n + 1) + Ω(5n^{2} + n, 5n + 2))

which establishes Theorem 1. Remark After submitting this manuscript, the author received a copy of [3] which contains a proof of an assertion equivalent to Theorem 1, using a similar method.

References

[1] H. M. Farkas & I. Kra, ‘On the quintuple product identity’, Proc. Amer. Math. Soc. 27 (1999), 771–778.

[2] D. Foata & G.-H. Han, ‘The triple, quintuple and septuple product identities revisited’, Séminaire Lotharingien de Combinatoire, B42o (1999), 12 pp.

[3] M. D. Hirschhorn, ‘An identity of Ramanujan, and applications’, Contemp. Math., to appear.