The Jacobi triple product identity is: $$F(z,q)=\prod\limits_{n=1}^{ \infty }(1-q^{2n})(1+zq^{2n-1})(1+z^{-1}q^{2n-1})=\sum\limits_{n = - \infty }^ \infty z^n q^{n^2} $$ where $|q|<1$
All roots of $F(z,q)=0$ for z can be expressed as:
$$z_k=-q^{2k-1}$$ where $k$ is an integer
I would like to find similar expansion for $\sum\limits_{n = - \infty }^ \infty n z^n q^{n^2} $ like The Jacobi triple product identity.
$$Q(z,q)=\sum\limits_{n = - \infty }^ \infty n z^n q^{n^2} $$
where $|q|<1$.
It is obvious that
$z=1$ and $z=-1$
$$Q(1,q)=Q(-1,q)=0$$
$z=1$ and $z=-1$ are trivial roots for $Q(z,q)=0$
Can we express roots ( $z_k=u_k(q)$) as known functions such as Theta functions , etc?
I would like to share my attempt to find $z_k=u_k(q)$: $$Q(z,q)=\sum\limits_{n = - \infty }^ \infty n z^n q^{n^2} $$ $$Q(z,q)=(z-z^{-1})q+2(z^2-z^{-2})q^4+3(z^3-z^{-3})q^9+4(z^4-z^{-4})q^{16}+5(z^5-z^{-5})q^{25}+.......$$
$$Q(z,q)=(z-z^{-1})q+2[(z-z^{-1})(z+z^{-1})]q^4+3[(z-z^{-1})(z^2+1+z^{-2})]q^9+4[(z-z^{-1})(z^3+z+z^{-1}+z^{-3})]q^{16}+5[(z-z^{-1})(z^4+z^2+1+z^{-2}+z^{-4})]q^{25}+.......$$
$$Q(z,q)=(z-z^{-1}) \big[ q+2(z+z^{-1})q^4+3[(z+z^{-1})^2-1)]q^9+4[(z+z^{-1})^3-2(z+z^{-1})]q^{16}+5[(z+z^{-1})^4-3(z+z^{-1})^2+1]q^{25}+.......\big]$$
We can easily see that trivial roots $z=1,-1$ can be gotten from $z-z^{-1}=0$
Other roots can be gotten from
$$Q(z,q)=(z-z^{-1})\big[(q-3q^9+5q^{25}+....)+(z+z^{-1})(2q^4-8q^{16}+...)+(z+z^{-1})^2(3q^9-15q^{25}+....)+(z+z^{-1})^3(4q^{16}+....)+(z+z^{-1})^4(5q^{25}+....)+...\big]$$
We can write that
$$Q(z,q)=(z-z^{-1})\big(a_0(q)+a_1(q)(z+z^{-1})+a_2(q)(z+z^{-1})^2+a_3(q)(z+z^{-1})^3+a_4(q)(z+z^{-1})^4+.....\big)$$
$$z+z^{-1}=T(q)$$
If $T(q)$ is root of $a_0(q)+a_1(q)T(q)+a_2(q)T(q)^2+a_3(q)T(q)^3+.....=0$
2 roots have a relationship :
$u_1=\frac{T(q)+\sqrt{T(q)^2-4}}{2}$;
$u_{-1}=\frac{T(q)-\sqrt{T(q)^2-4}}{2}$
$u_1=\frac{1}{u_{-1}}$
Some relations for $ Q(z,q) $ may also be helpful
$$Q(zq^2,q)=\sum\limits_{n = - \infty }^ \infty n z^n q^{n^2+2n} $$ $$zqQ(zq^2,q)=zq\sum\limits_{n = - \infty }^ \infty n z^n q^{n^2+2n} $$
$$zqQ(zq^2,q)=\sum\limits_{n = - \infty }^ \infty n z^{n+1} q^{n^2+2n+1} $$
$$zqQ(zq^2,q)=\sum\limits_{n = - \infty }^ \infty (n-1) z^{n} q^{n^2} $$ $$zqQ(zq^2,q)=\sum\limits_{n = - \infty }^ \infty n z^{n} q^{n^2} -\sum\limits_{n = - \infty }^ \infty z^{n} q^{n^2}$$
$$Q(z,q)-zqQ(zq^2,q)= \sum\limits_{n = - \infty }^ \infty z^{n} q^{n^2}$$
$$Q(zq^2,q)-zq^3Q(zq^4,q)= \sum\limits_{n = - \infty }^ \infty z^{n} q^{n^2+2n}$$ $$zqQ(zq^2,q)-z^2q^4Q(zq^4,q)= \sum\limits_{n = - \infty }^ \infty z^{n+1} q^{n^2+2n+1}$$
$$zqQ(zq^2,q)-z^2q^4Q(zq^4,q)=\sum\limits_{n = - \infty }^ \infty z^{n} q^{n^2}$$
$$Q(z,q)-zqQ(zq^2,q)=zqQ(zq^2,q)-z^2q^4Q(zq^4,q)$$
$$Q(z,q)+z^2q^4Q(zq^4,q)=2zqQ(zq^2,q) \tag{1}$$
Other relation can be written as: $$\frac{\partial F(z,q)}{\partial z}=\sum\limits_{n = - \infty }^ \infty n z^{n-1} q^{n^2}$$
$$Q(z,q)=z\frac{\partial F(z,q)}{\partial z} \tag{2}$$
Thank you for answers and comments
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EDIT:31/03/2020
Because $u_1, u_{-1}$ roots have relation :
$$u_1=\frac{1}{u_{-1}}$$
and if roots are $u_k(q)$ : where $k$ positive integer
We can write the expansion of $Q(z,q)$ as:
$$Q(z,q)=A(q) (z-z^{-1})\prod\limits_{k=1}^{ \infty }(1-\frac{z}{u_k(q)})(1-\frac{z^{-1}}{u_k(q)}) \tag{3}$$
The relation (3) satisfies $Q(z^{-1},q)=-Q(z,q)$
$A(q),u_k(q)$ only depend on $q$. I haven't found them yet . I have been looking for methods to find them.
Thanks for any helps to find them.
EDIT: 7th April 2020
I would like to add another relation . That can be helpful to find $A(q),u_k(q)$
$$F(z,q)=\prod\limits_{n=1}^{ \infty }(1-q^{2n})(1+zq^{2n-1})(1+z^{-1}q^{2n-1})$$
$$\ln F(z,q)=\ln \prod\limits_{n=1}^{ \infty }(1-q^{2n})+ \ln \prod\limits_{n=1}^{ \infty } (1+zq^{2n-1}) + \ln \prod\limits_{n=1}^{ \infty }(1+z^{-1}q^{2n-1})$$
$$\ln F(z,q)=\ln \prod\limits_{n=1}^{ \infty }(1-q^{2n})+ \sum\limits_{n = 1}^ \infty \ln(1+zq^{2n-1}) + \sum\limits_{n = 1}^ \infty \ln(1+z^{-1}q^{2n-1})$$
If we derivate both side for $z$;
$$\cfrac{\frac{\partial F(z,q)}{\partial z} }{F(z,q)}=\sum\limits_{n = 1}^ \infty \frac{q^{2n-1}}{1+zq^{2n-1}}-\sum\limits_{n = 1}^ \infty \frac{z^{-2}q^{2n-1}}{1+z^{-1}q^{2n-1}}$$
$$\cfrac{z\frac{\partial F(z,q)}{\partial z} }{F(z,q)}=\sum\limits_{n = 1}^ \infty \frac{zq^{2n-1}}{1+zq^{2n-1}}-\sum\limits_{n = 1}^ \infty \frac{z^{-1}q^{2n-1}}{1+z^{-1}q^{2n-1}}$$
$$\cfrac{Q(z,q)}{F(z,q)}=\sum\limits_{n = 1}^ \infty \frac{zq^{2n-1}}{1+zq^{2n-1}}-\sum\limits_{n = 1}^ \infty \frac{z^{-1}q^{2n-1}}{1+z^{-1}q^{2n-1}}$$
$$\cfrac{Q(z,q)}{F(z,q)}=\sum\limits_{n = 1}^ \infty \frac{zq^{2n-1}}{1+zq^{2n-1}}- \frac{z^{-1}q^{2n-1}}{1+z^{-1}q^{2n-1}}$$
$$\cfrac{Q(z,q)}{F(z,q)}=\sum\limits_{n = 1}^ \infty \frac{(z-z^{-1})q^{2n-1}}{1+q^{2(2n-1)}+q^{2n-1}(z+z^{-1})}$$
$$Q(z,q)=(z-z^{-1})F(z,q)\sum\limits_{n = 1}^ \infty \frac{q^{2n-1}}{1+q^{2(2n-1)}(1+\frac{q^{2n-1}}{1+q^{2(2n-1)}}(z+z^{-1}))}$$
$$Q(z,q)=(z-z^{-1})F(z,q)\sum\limits_{n = 1}^ \infty \frac{q^{2n-1}}{1+q^{2(2n-1)}}\big(1-\frac{q^{2n-1}}{1+q^{2(2n-1)}}(z+z^{-1})+\frac{q^{2(2n-1)}}{(1+q^{2(2n-1)})^2}(z+z^{-1})^2+.....\big]$$
$$Q(z,q)=(z-z^{-1})F(z,q)\sum\limits_{n = 1}^ \infty \frac{q^{2n-1}}{1+q^{2(2n-1)}}-\frac{q^{2(2n-1)}}{(1+q^{2(2n-1)})^2}(z+z^{-1})+\frac{q^{3(2n-1)}}{(1+q^{2(2n-1)})^3}(z+z^{-1})^2-.....\big)$$
$$Q(z,q)=(z-z^{-1})F(z,q)\sum\limits_{n = 1}^ \infty \sum\limits_{k = 0}^ \infty (-1)^k\frac{q^{(k+1)(2n-1)}}{(1+q^{2(2n-1)})^{k+1}}(z+z^{-1})^{k}$$
For $\Im(\tau)> 0$ let $P_\tau = \frac{-1-\tau}{2}+(0,1)+(0,\tau)$ be the fundamental parallelogram and (differientating as function of $z$) $$f(z;\tau)=\frac{\theta'(z;q)}{\theta(z;q)},\qquad \theta(z,q)= \sum_n e^{2i\pi nz} e^{i\pi n^2 \tau},\quad f(z+1;\tau)=f(z;\tau)$$ $$ f(z+\tau;\tau)=f(z;\tau)-2i\pi $$
The Jacobi triple product tells the zeros of $\theta(z;q)$ thus the poles of $f(z;\tau)$, it has one pole on $P_\tau$.
To find the number of zeros of $f(z;\tau)$ on $P_\tau$ look at $$\int_{\partial P_\tau} f(z;\tau)dz=2i\pi$$ Thus $f(z;\tau)$ has two zeros on $P_\tau$.
$\theta'(z;q)$ has a simple zero at $z=n$ and $z=n+1/2$. On the parallelograms $P_\tau+n+m\tau$ with $m \ne 0$ there are two zeros but their exact location is probably a special function of $\tau,m$, and not an elliptic integral.