# trigonometric → elliptic in the infinite product representation of θ3(z,q)

If $G(q) = \prod_{n=1}^{\infty} (1-q^{2n})$, then

$\theta_{3}(z,q) = G(q) \prod_{n=1}^{\infty} \left[1 - 2q^{2n-1}\cos(2z) +q^{4n-2}\right]$

Let’s define an analogue of the theta function $\theta_{3}(z,q)$ by replacing that trigonometric cosine $\cos$ with an elliptic $\mathrm{cn}(2z,k)$, where $k$ is the elliptic modulus.

So we’d have:

$A_{3}(z,q;k) = G(q) \prod_{n=1}^{\infty} \left[1 - 2q^{2n-1}\mathrm{cn}(2z;k) +q^{4n-2}\right]$

Here’s a domain colored picture of $A_{3}(z,1/2;1/2)$:

# another random sum, nothing fancy

$\sum_{n=2}^{\infty} [e^{\zeta(n)-1}-1] \approx 1.28715905135665420586\ldots$

# more morning speculation

since
$\int_{0}^{\infty} \frac{dx}{\Gamma(x^{3})} \approx 0.753065231886\ldots$

and
$\int_{0}^{\infty} \frac{dx}{\Gamma(x^{2})} \approx 1.165261151780\ldots$, then there is some number $\mu$ between three and two such that

$\int_{0}^{\infty} \frac{dx}{\Gamma(x^{\mu})} = 1$

# morning observation

$\sum_{n=1}^{\infty} \frac{1}{a^{n}} = \frac{1}{a-1}$

What my morning’s observation is that if we consider something like:
$\sum_{n=1}^{\infty} \frac{1}{2^{n}+n^{2}} \approx 0.5882390121819030646\ldots$

$\sum_{n=1}^{\infty} \frac{1}{3^{n}+n^{3}} \approx 0.3386248938590166285123\ldots$

$\sum_{n=1}^{\infty} \frac{1}{4^{n}+n^{4}} \approx 0.24096414317981403584949724\ldots$

$\sum_{n=1}^{\infty} \frac{1}{5^{n}+n^{5}} \approx 0.1877505985677791820123501\ldots$

$\sum_{n=1}^{\infty} \frac{1}{6^{n}+n^{6}} \approx 0.15415739781098121145677151253628\ldots$

So,:
$\sum_{n=1}^{\infty} \frac{1}{a^{n}+n^{a}} \propto 1/a$