Solve the non linear system of equations. We know that a function $f(t)$ has the  form 
$$
f(t)=x_1 e^{z_1 t}+x_2 e^{z_2 t}+\cdots+x_n e^{z_n t},
$$
for some unknowns $x_i, z_i$  but we can calculate the value  $f(t)$  for any $t.$
Question.  How to define the $x_1,x_2,\ldots, x_n, z_1, z_2, \ldots z_i$  in terms of the values $f(t)$?
My attempt. Put $y_i=e^{i z_1}.$  Then we get the following system of  non-linear equations
\begin{cases}
x_1+x_2+\cdots+x_n=f(0),\\
x_1 y_1+x_2 y_2+\cdots+x_n y_n=f(1),\\
x_1 y_1^2+x_2 y_2^2+\cdots+x_n y_n^2=f(2),\\
\ldots \\
x_1 y_1^{2n-1}+x_2 y_2^{2n-1}+\cdots+x_n y_n^{2n-1}=f(2n-1).
\end{cases}
Is there a nice general solution of  the system?
 A: Knowledge of theory of linear recurrences and university course of algebra [Kur] will guide us. 
Grouping equal exponents we can assume that $z_i$ are distinct and there are $m\le n$ of them. And, of course, all $x_i$’s are non-zero.  
Now for any natural $t$ function $f(t)$ satisfies the recurrence 
$$f(t+m)+a_{m-1} f(t+m-1)+\dots +a_0f(t+0)=0,$$
where $a_i$ are coefficients of the equation  
$$a(u)=y^m+a_{m-1}u^{m-1}+\dots +a_0=0,$$
whose roots are $u_i=e^{z_i}$, $i=1,\dots, m$.
So first we solve a system $\bar f+Fa=0$ of $m$ linear equations to find $a_i$’s. Here $$\bar f=(f(m),f(f+1),\dots, f(2m-1))^T,$$ $F=\|f_{ij}\|,$ where $i,j=1..m$ and for each $i,j$ $$f_{ij}=f(i+j-2)=\sum_{k=1}^m x_ku^{i+j-2}_k.$$  Thus $F=GH$, where $G=\|g_{ij}\|$ and $H=\|h_{ij}\|$ are $m\times m$ matrices, and $g_{ij}=u_j^{i-1}$, $h_{ij}=x_iu_i^{j-1}$, for each $i,j$. Thus $$\det F=\det G\cdot\det H=(\det G)^2 \cdot x_1\cdots x_m.$$ But $\det G=\prod_{i<j} (u_j-u_i)$, because it is Vandermonde determinant. (By the way, $(\det G)^2=(-1)^{\frac {m(m-1)}2}R(a,a’)$ is a discriminant of the polynomial $a$, and $R(a,a’)$ is a  resultant of the polynomial $a$ and its derivative $a’$ (see, for instance, [Kur, p.343-345]). Thus  $$\det F=\prod_{i<j} (u_j-u_j)^2 x_1\cdots x_m\ne 0.$$ Hence the system $\bar f+Fa=0$ has a unique solution $a$. Next we solve a polynomial equation $a(u)=0$ to find $u_i$’s. Finally, we again solve a  system $Gx=\hat f$ of $m$ linear equations to find $x_i$’s. Here  $$\hat f=(f(0),f(1),\dots, f(m-1))^T.$$ Since $\det G\ne 0$, this system has a unique solution. 
References
[Kur] Kurosh A.G., “Course of the highest algebra”, Moskow, Nauka, 1968, 9-th edition (in Russian).
