# What can determine $a_k$ and $\omega_k$ in $t \mapsto \sum_{k=1}^n a_k e^{i\omega_kt}$?

If $$p$$ is a polynomial in $$t$$ of degree $$n$$, say $$p(t) = a_nt^n + \cdots + a_1t + a_0,$$ then if we choose $$n+1$$ distinct points $$t_0, \dots, t_n$$, the $$a_k$$s are completely determined by the values of $$p(t_0), \dots, p(t_n)$$.

If we define $$c(t) = a_0e^{i\omega_0t} + \cdots + a_ne^{i\omega_nt},$$ can we determine the $$a_k$$s and $$\omega_k$$s by looking at the values of $$c$$ on finitely many choices of $$t$$?

$$c(t) = \sum_{m=1}^n a_m e^{i w_m t}$$ the $$w_m$$ are distinct and the $$a_m$$ are non-zero.
The shifts of the functions $$e^{i w_m t}$$ generate a vector space of dimension $$n$$, the same as the vector space generated by the shifts of $$c(t)$$.
Pick some $$t_1,\ldots,t_n$$ such that the $$c(t+t_j)$$ are linearly independent, we get some $$B_{m,j}$$ such that $$\forall m, \forall t, \qquad e^{i \omega_m t} = \sum_{j=1}^n B_{m,j} c(t+t_j)$$
from there we have the linear maps $$\pmatrix{c(t+T+t_1)\\ \vdots \\ c(t+T+t_n)} = B^{-1} e^{i \text{ diag}(w) T} B \pmatrix{c(t+t_1)\\ \vdots \\ c(t+t_n)}$$
Your question reduces to sample finitely many values of $$c$$ to recover $$B^{-1} e^{i \text{ diag}(w)} B$$ and $$B^{-1} e^{i \text{diag}(w) T} B$$ with say $$T=\pi$$, diagonalize those matrices to recover $$i w \bmod 2i \pi$$ and $$i w T\bmod 2i \pi$$ from which we know $$w$$ and hence $$a$$.