Relationship between Fourier coefficients of $f\left(x\right)$ and $f^{-1}\left(x\right)$

Question

Say I have a function $f\left(x\right)$, which can be expressed as a Fourier Series:

$$f\left(x\right)=\sum_{k=-\infty}^{\infty} c_k e^{ikx}$$

Define the inverse of $f\left(x\right)$ as, $f^{-1}\left(x\right)=1/f\left(x\right)$, with its Fourier Series:

$$f^{-1}\left(x\right)=\sum_{k=-\infty}^{\infty} {c'}_k e^{ikx}$$

Is there a known relationship between ${c'}_k$ and $c_k$? I.e. if I know $c_k$, can I get ${c'}_k$ directly?

For my specific case I am interested in functions for which the following are true:

1. $x\in\left[0,2\pi\right]$
2. $0 \lt f\left(x\right)$
3. $f\left(x\right)$

Answer 1

It appears that, at least under certain conditions, the coefficients of $1/f$ can be obtained from the coefficients of $f$. Per the last comment on THIS thread, see the following two papers:

"A. Edrei and G. Szegö, “A note on the reciprocal of a Fourier series,” Proc. Am. Math. Soc., vol. 4, no. 2, pp. 323–329, 1953."

"R. Duffin, “The Reciprocal of a Fourier Series,” Proc. Am. Math. Soc., vol. 13, no. 6, pp. 965–970, 1962."

Answer 2

The relationship between the coefficients of $f$ is $1/f$ is opaque. Consider the special case when $\hat f\in \ell^1(\mathbb Z)$ (hence, the function $f$ is continuous). If $f$ does not vanish, then the Fourier coefficients of $1/f$ are also in $\ell^1$; this is Wiener's $1/f$ theorem. If you could get the coefficients of $1/f$ directly, you'd have a constructive proof of Wiener's theorem, with some explicit $\ell^1$ norm estimate. But there is not really such a thing, despite some partial results. See the paper In search of the invisible spectrum by Nikolskii. I remember attending the author's talk titled "Why there exists no constructive proof of Wiener’s $1/f$ theorem?" (as seen here) but, sadly, no longer remember the reason why.