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How can I prove the following?

$$1-x+x^2+x^5-x^7-x^{12}+x^{15}-x^{22}-x^{26}+x^{35}-x^{40}+\dots \\= \prod_{i=1}^{\infty} [(1 - x^{8 i - 7}) (1 + x^{8 i - 6}) (1 + x^{8 i - 5}) (1 + x^{8 i - 4}) (1 + x^{8 i - 3}) (1 + x^{8 i - 2}) (1 - x^{8 i - 1}) (1 - x^{8 i})]$$

It doesn't seem to follow from the Triple Product formula and I haven't been able to come up with a combinatorial proof.

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  • $\begingroup$ What is that left hand side? Is there a pattern to the signs? Or are you just claiming that the coefficients are all $\pm 1$ (or $0$)? $\endgroup$
    – Zach Teitler
    Commented Jul 14, 2017 at 17:30
  • $\begingroup$ @ZachTeitler It is the series for the expansion of $\prod_i(1-x^i)$ but with different signs. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Jul 14, 2017 at 17:31
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    $\begingroup$ Coefficients of lhs seem to be A206958 in OEIS. If so then according to that page, it is the expansion of$f(x^5,-x^7) -xf(-x,x^{11})$, where$$f(a,b)=1 +(a+b) +(ab)(a^2+b^2) +(ab)^3(a^3+b^3) +(ab)^6(a^4+b^4) + (ab)^{10}(a^5+b^5)+...$$is the Ramanujan theta function with the infinite product representation $(-a;ab)_\infty(-b;ab)_\infty(ab;ab)_\infty$ $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Jul 14, 2017 at 18:36
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    $\begingroup$ @მამუკაჯიბლაძე Yes, that identity follows from the Riemann addition relation of theta functions, see relation (5) in mathworld.wolfram.com/QuintupleProductIdentity.html $\endgroup$
    – Gjergji Zaimi
    Commented Jul 14, 2017 at 19:02

1 Answer 1

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This is an instance of Watson's quintuple product identity (also Macdonald identity for $BC_1$): $$\prod_{n\geq 1}(1-s^n)(1-s^nt)(1-s^{n-1}t^{-1})(1-s^{2n-1}t^2)(1-s^{2n-1}t^{-2})=\sum_{n\in \mathbb Z}s^{\frac{3n^2+n}{2}}(t^{3n}-t^{-3n-1}).$$ By plugging in $t=x^{-1}$ and $s=-x^{4}$ this becomes exactly your identity.

Somewhat amusingly, the quintuple product identity can be proven directly from the triple product identity. See this article by Carlitz, or this article of Foata and Han.

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