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Consider a complex Fourier series $$f(\varphi)=\sum_{m=-\infty}^\infty a_me^{im\varphi}$$ Its reciprocal (where it exists) also admits a Fourier series expansion $$\frac{1}{f(\varphi)}=\sum_{m=-\infty}^\infty b_me^{im\varphi}$$ I would like to obtain an explicit formula for the coefficients $b_m$ in terms of the coefficients $a_m$, ideally as some kind of infinite sum over (products of) the coefficients. Does anyone know of such a formula?

Possible simplification

In practice, I am most interested in the zero-mode $b_0$, and I am completely happy to assume that only finitely many modes $a_m$ are nonzero (i.e., that there exists some $k$ such that $a_m=0$ for $|m|>k$).

A possible representation of the answer

I have only found two papers in the literature tackling this question. The first is a 1953 paper by Edrei and Szego, and the other is a 1962 paper by Duffin:

https://www.jstor.org/stable/2031811

https://www.jstor.org/stable/2034097

Equation (4) in the Edrei-Szego paper (visible without going through the paywall) gives a formula for the $b_m$ when $f(\varphi)$ is a real, positive function. In that case, they point out that the problem is equivalent to the solution of a linear Toeplitz system, where the infinite Toeplitz matrix is Hermitian because $f$ is real. (Moreover, note that the matrix has narrow bandwidth $k$ if $a_m=0$ for $|m|>k$.)

That equation is not explicit since one must still compute some ratio of determinants. However, it takes the form of the limit (as the the $n\times n$ Toeplitz matrix $T_n$ grows to infinite size) of the ratio of the determinant of the leading principal minor of $T_n$ to the determinant of $T_n$.

In the 2002 paper on "Toeplitz Minors" by Bump and Diaconis (https://www.sciencedirect.com/science/article/pii/S0097316501932145), they seem to provide (on the second page) a formula for precisely such a ratio in the limit $n\to\infty$. I'm a little out of my depth here---can anyone confirm that this is indeed what they are computing? If so, does their general formula give an explicit formula for the case that I'm interested in here? (Their formula seems to also hold for more complicated principal minors in the numerator of the ratio.)

Or if this is not a helpful path to go down, has anyone encountered any other explicit way of representing $b_0$ as an explicit infinite sum over some combination of the $a_m$?

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An algorithm for this purpose has been propsosed by Alan Guida, On the Reciprocal of a Fourier Series (1978). The algorithm is most efficient if the Fourier series is given in the factored form $f(\phi)=\prod_{n=1}^N(e^{i\phi}-c_n)$.


If you only wish to obtain the zero-mode $b_0$, and you have the factored form of $f(\phi)$, then you can obtain it directly from the integral $$b_0=\frac{1}{2\pi i}\oint_{|z|=1}\frac{dz}{z}\prod_{n=1}^N(z-c_n)^{-1}=\prod_{n=1}^N(-c_n)^{-1}+\sum_{|c_n|<1}\frac{1}{c_n}\prod_{n'\neq n}(c_n-c_{n'})^{-1}.$$

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  • $\begingroup$ Thank you, Carlo, both for the reference and for working out that formula for $b_0$. Unfortunately as soon as the factored form contains even a moderate number of terms, it becomes impossible to represent the $c_n$ in terms of the $a_n$. Do you think doing so is simply not possible? $\endgroup$ Commented Aug 31, 2023 at 14:03
  • $\begingroup$ the $c_n$'s are roots as a function of $z=e^{i\phi}$ of the Fourier series $\sum a_m z^m$; a method to find these is discussed at math.stackexchange.com/a/372166/87355 $\endgroup$ Commented Aug 31, 2023 at 16:01
  • $\begingroup$ The formula in your post is very useful if the Fourier series is given in factored form, but when is that the case? If one is given only the Fourier modes $a_n$ (as is usually the case), then solving for the roots $c_n$ in closed form is not possible. One might as well solve the matrix problem in Duffin's paper. $\endgroup$ Commented Sep 2, 2023 at 21:50
  • $\begingroup$ thank you, hope to be of more help another time! $\endgroup$ Commented Sep 4, 2023 at 20:05

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