# Inverse Fourier transform of $F(\omega)=\frac{1}{\sum_{k=1}^N e^{i \omega z_k}}$

I am looking for the inverse Fourier transform of \begin{align*} F(\omega)=\frac{1}{\sum_{k=1}^N e^{i \omega z_k}} \end{align*} where $z_k \in Z$. But I don't know how to approach it. This reminds me of train of delta function. For example Fourier transform of $\mathcal{F}^{-1}(\frac{1}{ e^{i \omega z_k}})= \delta(t-z_k)$

In general the inverse Fourier transform is not well defined. Observe that for generic choices of $z_k$ there exists $\omega$ such that the denominator of your expression vanishes. This implies that $F(\omega)$ is not even a tempered distribution, so you can't really even define its (inverse) Fourier transform.

Supposing the inverse Fourier transform is defined, then we expect that it would be a function/distribution $f$ verifying $$\delta(t) = \sum_{k = 1}^N f(t - z_k)$$ This morally gives an infinite system of linear equations. In the case where $z_k$ are not commensurate, this gives rise to an uncountably infinite system of linear equations; this case is hard to deal with. In the case where $z_k$ are commensurate, we can rescale and consider the equivalent problem where $z_k\in \mathbb{Z}$. Then in principle we reduce this to a countably infinite system of finite linear equations. In this case we can frequently approach this by generating functions or recursive relations. But note that the solution may or may not be unique.

For example, take $z_1 = 0$ and $z_2 = 1$ and $N = 2$. We can write $f(t) = \sum_{j \in \mathbb{Z}} a_j \delta(t - j)$. Then we have $$a_j + a_{j-1} = 0, j\neq 0$$ and $$a_0 + a_{-1} = 1$$ For this system it is easy to see that for each choice of $a_0$ there exists a solution. None of the solutions, however, can verify $\lim_{|j|\to\infty} = 0$. (This is of course related to the fact that when $z_1 = 0$ and $z_2 = 1$ we have that when $\omega = (2k+1)$ for $k\in\mathbb{Z}$, we have that $F(\omega) = \infty$.)

On the other hand, if we take $N = 3$ and $z_1 = z_2 = 0$ and $z_3 = 1$, the equivalent system becomes $$2 a_j + a_{j-1} = 0, j\neq 0$$ and $$2 a_0 + a_{-1} = 1$$ Even though it is still the case that for every value of $a_0$ there exists a solution, there is an exceptional one where $$a_0 = 1/2,$$ $$a_{j} = 0, j < 0$$ $$a_j = (-1)^j / 2^{j+1}, j > 0$$ which has the property that the coefficients decay to zero as $|j|\to\infty$. The existence of this solution is related to the fact that $2 + e^{i\omega} \neq 0$.

(The commensurate case implies that $F(\omega)$ is periodic, and so the problem can be reduced to finding its Fourier series; the coefficients $a_j$ above are basically the Fourier coefficients, at least in the case when $F(\omega)$ remains bounded. For more information on solving countable systems of finite linear equations, you can also look at this article and its references.)

• Thank you. This an impressive answer. I have few questions for you if you don't mind. My first question is what does it mean for $z_k$ to commensurate? – Boby Jan 22 '15 at 3:50
• meaning that for any $j,k$ you have $z_k/z_j$ a rational number. The $z_j$ themselves don't have to be rational. For example, $\pi$ and $3\pi/4$ are commensurate; but $\sqrt{2}$ and $\sqrt{3}$ are not. – Willie Wong Jan 22 '15 at 4:30
• Thanks. My second question. Would it be easier if I told you that $\frac{1}{N}\sum_{k=1}^N z^2_k=1$ ? – Boby Jan 22 '15 at 15:40
• In general: no. Take $N = 3$. You can have $0,1, \sqrt{2}$, or $-\sqrt{3/2}, \sqrt{3/2}, 0$, or $1,1,1$ which are all good/bad for different reasons. But maybe it can help someone else with a more detailed and better answer. – Willie Wong Jan 23 '15 at 2:17