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During my research on extension of the Jacobi Triple Product. , I finally got an interesting conjecture that is similar and extension of of the Jacobi Triple Product for higher terms . I asked a related question 3 years ago. (The link of my question is) . It should be true with strong sense. I need help to disprove or prove my conjecture (1) below. I do not know if it is known theorem or not.

$$[\phi(qh)-\phi(qh^{-1})]\prod\limits_{n=1}^{ \infty }(1-q^{n}h^{n^2})(1-q^{n}h^{-n^2})=\sum\limits_{n = - \infty }^ \infty (-1)^n q^{\frac{n(n+1)}{2}} h^{\frac{n(n+1)(2n+1)}{6}} \tag 1 $$

Where $\phi(q)$ is Euler function and $h=e^{i.x}$ , $x$ is Real Number

$$\phi(q)=\prod\limits_{n=1}^{ \infty }(1-q^{n})=\sum\limits_{n = - \infty }^ \infty (-1)^n q^{\frac{n(3n-1)}{2}} \tag 2$$

if the series are extended in positive powers of $q$ with coefficients that are Laurent polynomials in $h$, I confirmed the Equation (1) for first few terms (I have confirmed $q^3$ terms manually for now) . I do not have math software extend such q series with $h$ variables in it .

Could you please help me to confirm more terms if my conjecture is true for more terms? I will be very appreciated for confirmation for higher terms or disprove for higher terms that the conjecture is false for higher terms.

$$\phi(qh)-\phi(qh^{-1})= q(h^{-1}-h)+q^2(h^{-2}-h^{2})-q^5(h^{-5}-h^{5})+.... \tag 3 $$

$$\prod\limits_{n=1}^{ \infty }(1-q^{n}h^{n^2})(1-q^{n}h^{-n^2})=1-q(h^{-1}+h)+q^2(1-h^{-4}-h^4)+q^3(h^3+h^{-3}+h^5+h^{-5}-h^9-h^{-9})+....\tag 4$$

$$[\phi(qh)-\phi(qh^{-1})]\prod\limits_{n=1}^{ \infty }(1-q^{n}h^{n^2})(1-q^{n}h^{-n^2})=q(h^{-1}-h)-q^3(h^{-5}-h^5)+.... \tag 5$$

$$\sum\limits_{n = - \infty }^ \infty (-1)^n q^{\frac{n(n+1)}{2}} h^{\frac{n(n+1)(2n+1)}{6}}=q(h^{-1}-h)-q^3(h^{-5}-h^5)+.....\tag 6$$

Thanks a lot for helps

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  • $\begingroup$ When $h=1$, the LHS is zero but the RHS is not. Did you miss a "$+1$" on the LHS? My suspicion is that generalizing Jacobi is unlikely. $\endgroup$ – T. Amdeberhan Dec 28 '16 at 16:29
  • $\begingroup$ @T.Amdeberhan When $h=1$ , $\sum\limits_{n = - \infty }^ \infty (-1)^n q^{\frac{n(n+1)}{2}}=0$. It is not problem. $\endgroup$ – Mathlover Dec 28 '16 at 16:35
  • $\begingroup$ Oh, you're right. $\endgroup$ – T. Amdeberhan Dec 28 '16 at 16:45
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Your conjecture is false. The coefficient of $q^{4}$ in the left hand side is $$ -h^{-10} + h^{-8} + h^{-2} - h^{2} - h^{8} + h^{10} $$ (and not zero, like you had conjectured).

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  • $\begingroup$ Thanks for answer. I will double check my all notes again . I will update my conjecture. $\endgroup$ – Mathlover Dec 28 '16 at 16:37

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