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I have a system of equations $$ S_{m}(\xi) +P_{m}(\xi)=f(\xi) $$ where $\xi$ can be choosen arbitrary in some domain in $\mathbb{C}$, $f$ is known, $P_m$ is a polynomial of degree at most $m$. Here $S_{m}$ is a generic function from $\mathbb{C}$ to $\mathbb{C}$ but for me it is sufficient to know it's values in only $(m+1)$ mutually distinct points.

Taking $(m+1)$ mutually distinct points $\xi_0, \ldots, \xi_m$ we can invert the Vandermonde matrix and hence write coefficients of $P_m$ as linear combination of $S_m(\xi_k)$,$k=0,\ldots,m$. Introducing this result back in equations we get a linear system on $S_{m}(\xi_k)$. So theoretically we can find $S_{m}(\xi_k)$. But is there some way to do it practically?

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Solving the system for $P$ and then inserting the result back into the same system will just give you a tautology ($x=x$). – David Ketcheson Aug 13 '12 at 12:03
@DavidKetcheson No, I can just add other points – Nimza Aug 13 '12 at 12:05
What are your unknowns here? It sounds like you're actually trying to build up a polynomial approximation $P_m$ of $S_m$ by sampling values $\xi$, then using that polynomial fit to estimate the values of $S_m$ at other points. – Aron Ahmadia Aug 13 '12 at 13:29

3 Answers

If you know $f$ and $S$ in $m+1$ points, you have the standard polynomial interpolation problem with $f-S$ as right hand side. Thus newton or lagrange interpolation give stable answers.

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Unfortunately I don't know both $S_n$, $P_n$. So I can't separate $S_n$ from $P_n$. – Nimza Aug 13 '12 at 15:50
So you don't need $S_n$, just the evaluation of $S_n$ at the $m+1$ points. – Aron Ahmadia Aug 13 '12 at 16:06
@Nimza: You need $f$ ans $S$ just at $m+1$ points to find $P$. – Arnold Neumaier Aug 13 '12 at 18:02

Unfortunately I can't comment yet, so I'm putting this as an answer.

I don't think there are enough constraints in the problem. $S_m = f - P_m$ where $P_m$ is any polynomial of degree $\leq m$ is going to be a possible solution, so there isn't going to be a way to find out which polynomial we're talking about.

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There are far better polynomial basis to use than the Vandermonde, which turns out to be really poorly conditioned for a lot of real-world fits. If you know something about your solution, you'll be able to pick a better basis, but I would start with the Chebyshev basis appropriately scaled by the range of $\xi$.

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