What can be said about the solutions of, resp. solving, the following system polynomial equations, in which the $x_i$ and $y_j$ are the variables and $c_{i,j},\,d_i\in\mathbb{R}$ are constants:
$$\begin{matrix}
c_{0,1}x_1+\,\cdots+\,c_{0,n}x_n&=&d_0\\
c_{1,1}x_1y_1+\,\cdots+\,c_{1,n}x_ny_n&=&d_1\\
c_{2,1}x_1y_1^2+\,\cdots+\,c_{2,n}x_ny_n^2&=&d_2\\
\vdots \\
c_{i,1}x_1y_1^i+\,\cdots+\,c_{i,n}x_ny_n^i&=&d_i\\
\vdots\\
c_{m,1}x_1y_1^{m}+\,\dots+\,c_{m,n}x_ny_n^{m}&=&d_{m}
\end{matrix} $$
knowing the conditions and algorithms for efficient numeric or symbolic calculation of the solutions would be the precondtion for an algorithm for non-polynomial splines as requested in this MO question Non-polynomial splines, a non-linear problem
1 Answer
For the special case where all $c_{i,j}$'s are equal to 1 and $m=2n-1$, take a look at Ramanujam's paper: http://ramanujan.sirinudi.org/Volumes/published/ram03.pdf. Needless to say, it is an ingenious method. The steps are the following:
(1) The key idea is to recognize that the coefficients (w.r.t $\theta$, upto the $2n-1$ order) of $\sum_{k=1}^n\frac{x_k}{1-\theta y_k}$ in the series expansion would be the LHS of the nonlinear equations.
(2) Now, $\sum_{k=1}^n\frac{x_k}{1-\theta y_k} = d_1 + d_2\theta + \cdots + d_{2n}\theta^{2n-1} + \cdots = \frac{A_1 + A_2\theta + \cdots + A_{n-1}\theta^{n-1}}{B_1 + B_2\theta + \cdots + B_{n-1}\theta^{n-1}}$.
(3) Multiplying by $B_1 + B_2\theta + \cdots + B_{n-1}\theta^{n-1}$ on either side and comparing coefficients, gives linear equations in $A_i$'s and $B_i$'s.
(4) If a solution exists, one can then determine partial fractions (in $\theta$), and the coefficients would be the answers.
For your case, you might want to think on the same lines with Pade approximants.
All this is only a suggestion for the most general case. Hope it helps.
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1$\begingroup$ the linked paper seems to be the door-opener and appetizer for tackling the more general problem; thank you very much for unearthing the reference! $\endgroup$ Commented Apr 19, 2020 at 16:38
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$\begingroup$ The link in this answer seems to be broken now... $\endgroup$ Commented Apr 10, 2022 at 15:59
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$\begingroup$ this coding.yonsei.ac.kr/ramanujan_paper.pdf is an alternative link to Ramanujan's paper $\endgroup$ Commented Aug 17, 2022 at 9:28