# How can i reconstruct a polynomial with access to its roots and not to some points evaluation of it?

Is there any other element that can let me interpolate a polynomial other than the well known lagrange interpolation by which a polynomial of degree $d$ can be reconstructed from $d+1$ pairs: $x_i,f(x_i)$. If i have another kind of information which is related with the polynomial, ie: some roots can i reconstruct it?

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Aren't roots also points of evaluation: $(x_i,0)$? – draks ... Oct 9 '12 at 8:22
right. But is there something special on the evaluation on 0 that makes it sufficient to reconstruct it with less than d+1 points? – curious Oct 9 '12 at 8:23

## 2 Answers

If you know all the roots of a polynomial, then you can almost reconstruct it. If $a_1,\dots,a_n$ are the roots, then the polynomials with exactly those roots are of the form $c(x-a_1)(x-a_2)\cdots (x-a_n)$ for some constant $c$ (assuming we really know all roots, ie that we are not necessarily limiting ourselves to the real roots or something like that).

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To be pricise, this answer requires that you know all the roots and their multiplicities, and that this includes roots that lie in an extension of the ground field (so for a real polynomial, you need to know the complex roots, and their multiplicities, as well). – Marc van Leeuwen Oct 9 '12 at 9:23

I believe the polynomial is uniquely characterized by its coefficients, from which in an order n case you will have n+1 . Identifying the polynomial on which a set of sampling points lie thus corresponds to the estimation of its coefficients that can be achieved by solving a linear system of equations (polynomial regression). In order for the system to have a unique solution the system needs to have a rank of n+1 in case of a degree n polynomial (the matrix may not be square assuming the presence of redundant rows as a result of an overdetermined system, assuming no noise having more than n+1 points results in redundancies). For this reason in the general case you need at least n+1 different points to be able to find the solution.

Peter

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