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Problem Statement

I am given a finite set of monic polynomials in t, parameterized by $r_i$

$X_i = t - r_i$

where the $r_i$ are guaranteed unique. Neither $t$ nor $r_i$ are known, only $X_i$. I would like to solve for t or every $r_i$.

Attempt at solution using elementary polynomials.

I am trying to use elementary symmetric polynomials to solve for $t$. I know that

$g_n(t+r_i) = \prod_j^n X_j = \sum_j^{n} (-1)^j \sigma_{j,n} (t^{n-j})$

where $\sigma_{j,n}$ is the j-th elementary symmetric polynomial on n variables (usually this is written just $\sigma_{j}$).

Because I don't have access to the roots themselves, I am trying something like:

$\Delta_n = g_{n+1} - g_{n} = (t - r_{n+1} -1)\prod_{i}^{n}(X_i) = \sum_j^{n+1} (-1)^j \sigma_{j,n+1} (t^{n-j+1}) - \sum_j^{n} (-1)^j \sigma_{j,n} (t^{n-j})\\ = [(-1)^{n+1}\sigma_j^{n+1}] + t^n \sum_j (-1)^j t^{-j} [t \sigma_{j,n+1} - \sigma_{j,n}] $

I would like to somehow relate $\sigma_{j,n+1}$ to $\sigma_{j,n}$, with the intention of simplifying the bracketed last term and solving for $t^n$ using the known product $(t - r_{n+1} -1)\prod_{i}^{n}(X_i)$. Is such a thing possible? Barring that, are there other methods to solve the system? The closest thing I can find on this site relates $\sigma_{j+1,n}$ to $\sigma_{j,n}$. I have also looked at methods for approximating the roots using derivatives and exact solutions given knowledge of the roots (pp5)

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heisenBug
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1 Answers1

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$t=0$, $r_i=-X_i$ is a solution. For any constant $c$, $t=c$, $r_i=c-X_i$ is a solution.

Adding higher order complications will not change anything in this picture. Is there some information missing?

Lutz Lehmann
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  • The information missing is that I am working in a system in which I know $X_i = t - r_i$, but neither t not $r_i$ are individually known, controllable, or separable. I only have $X_i$. If it were otherwise, you would be absolutely correct. – heisenBug Jun 05 '14 at 02:24
  • You already said that. Your system of equations is still underdetermined, $n$ equations for $n+1$ unknowns. This gives a solution set of dimension 1, in this case a straight line as parametrized in the answer. – Lutz Lehmann Jun 05 '14 at 08:17