Prove that $\sum_{n=0}^\infty\frac{x^{2n+1}}{1+x^{4n+2}}\neq0$ for complex $x$

Question

For $x\in\mathbb C$ with $0<|x|<1$, does this function have any roots?

$$f(x)=\sum_{n=0}^\infty\frac{x^{2n+1}}{1+x^{4n+2}}$$

$$=\sum_{n=0}^\infty x^{2n+1}\sum_{m=0}^\infty(-x^{4n+2})^m$$

$$=\sum_{n=0}^\infty\sum_{m=0}^\infty(-1)^mx^{(2n+1)(2m+1)}$$

Finding an infinite product representation would help. I don't remember where I found this, but here's another form of the function:

$$f(x)\overset?=\sum_{n=0}^\infty\sum_{m=0}^\infty x^{((4n+1)^2+(4m+1)^2)/2}$$

Answer 1

A correction to the final formula, and a product representation (one of): $$f(x)=\sum_{n,m=0}^{\infty}x^{((\color{red}{2}n+1)^2+(\color{red}{2}m+1)^2)/2}=x\prod_{n=1}^{\infty}(1-x^{4n})^{2(-1)^n}.$$ As noted in the comments to the OP, this has much to do with theta functions. In that notation, $$f(x)=\frac{\theta_4'(\pi/4,x)}{4\theta_4(\pi/4,x)}=\frac{\theta_1'(0,x^4)}{2\theta_4(0,x^4)}=\frac{\theta_2^2(0,x^2)}{4},$$ where $\theta_k'(z,q)=\partial\theta_k(z,q)/\partial z$. Here, the first equality is taken from the logarithmic derivative of the product representation for $\theta_4(z,x)$, at $z=\pi/4$; the second one follows from the defining series; the third one is obtained again using the product representations.

Answer 2

The last form, as corrected by @metamorphy, follows from this fact: Given that an exponent $N$ on $x^{1/2}$ is congruent to $2\text{ mod }4$, the number of ways $N$ can be written as a sum of two squares (which are necessarily odd) is equal to the difference between the numbers of factors of $N$ congruent to $1$ and $3\text{ mod }4$. (Mathologer and 3Blue1Brown made videos on this, but the proof is not self-contained, as it relies on uniqueness of Gaussian prime factorization.) Then $N$ can be written as $2$ times that factor times another odd number, so the form $(-1)^mx^{(2m+1)(2n+1)}$ appears.

Continuing from there, we have

$$f(x)=\sum_{n=0}^\infty\sum_{m=0}^\infty x^{((2n+1)^2+(2m+1)^2)/2}$$

$$=\sum_{n=0}^\infty\sum_{m=0}^\infty x^{2n(n+1)+2m(m+1)+1}$$

$$=x\left(\sum_{n=0}^\infty x^{2n(n+1)}\right)^2$$

$$=x\left(\frac12\sum_{n\in\mathbb Z}x^{2n(n+1)}\right)^2$$

$$=x\left(\frac12\sum_{n\in\mathbb Z}(x^2)^n(x^2)^{n^2}\right)^2$$

and the Jacobi triple product formula gives

$$=x\left(\frac12\prod_{n=1}^\infty\big(1-x^{4n}\big)\big(1+x^{4n}\big)\big(1+x^{4n-4}\big)\right)^2$$

$$=x\left(\prod_{n=1}^\infty\big(1-x^{4n}\big)\big(1+x^{4n}\big)^2\right)^2.$$

This is sufficient, but I'll show its equivalence to the other answer:

$$=x\left(\prod_n\big(1-x^{8n}\big)\big(1+x^{4n}\big)\right)^2$$

$$=x\left(\prod_n\frac{\big(1-x^{8n}\big)\big(1-x^{8n}\big)}{\big(1-x^{4n}\big)}\right)^2$$

$$=x\left(\frac{\prod_{\text{even }n}\big(1-x^{4n}\big)\prod_{\text{even }n}\big(1-x^{4n}\big)}{\prod_{\text{odd }n}\big(1-x^{4n}\big)\prod_{\text{even }n}\big(1-x^{4n}\big)}\right)^2$$

$$=x\frac{\prod_{\text{even }n}\big(1-x^{4n}\big)^2}{\prod_{\text{odd }n}\big(1-x^{4n}\big)^2}.$$

Answer 3

Just to mention that we are dealing with modular forms.

Using that $\frac{\pi}{\cosh(\pi t)}$ is its own Fourier transform and the Poisson summation formula we get that $$f(z)=\sum_{n=0}^\infty\frac{e^{2i\pi (2n+1) z}}{1+e^{2i\pi (4n+2) z}}$$ is a weight $1$ modular form for $\Gamma_1(8)$,

then $$F(z)=\prod_{\gamma \in \Gamma_1(8)\setminus SL_2(Z)} f(\gamma(z))(cz+d)^{-1}$$ is a weight $[SL_2(Z): \Gamma_1(8)]=48$ cusp form for $SL_2(Z)$. We find that $F(z)=O(e^{8i\pi z})$ at $i\infty$, thus $F(z)/\Delta(z)^4$ is a weight $0$ modular form, it has to be constant.

$\Delta(z)$ is given by an infinite product, thus $F$ doesn't vanish, $f$ doesn't vanish.