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Given two factored polynomials of the same degree $N$:

$$ \begin{align} P(x) &= \prod_{k=1}^{k=N} (x - p_k) \\ Q(x) &= \prod_{k=1}^{k=N} (x - q_k) \end{align} $$

Due to $P$ and $Q$ having the same degree there exists a poylnomial $P'$ of degree $N - 1$ and number $C$ (i.e. polynomial of degree zero) such that:

$$ C + \frac {P'(x)}{Q(x)} = \frac {P(x)} {Q(x)} $$

Where $P'$ is needed in factored form also (a constant $K$ is introduced as leading coefficient may not be 1):

$$ P'(x) = K \prod_{k=1}^{k=N - 1} (x - p'_k) $$

Is it possible to find $P'$ in factored form given both $P$ and $Q$ are in factored form?

Has there been any work done on a numerically stable computer algorithm to do this?

One method would be to compute the partial fraction expansion of $\frac {P} {Q}$ and do some heavy rearranging and then find the roots of $P'$ but that is computationally expensive and will lead to inaccuracies for large $N$.

It can be assumed the $P$ and $Q$ have no repeated roots and no roots in common and that in their leading coefficient is 1 so that:

$$ \begin{align} P(x) &= 1 + a_0 x + ... + a_N x^N \\ Q(x) &= 1 + b_0 x + ... + b_N x^N \\ P'(x) &= K(1 + c_0 x + ... + c_{N-1} x^{N - 1}) \end{align} $$

keith
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  • P and Q are assume with principal coefficient 1? Why the derivate must have the same property? – Martín Vacas Vignolo May 24 '16 at 11:50
  • @vvnitram, yes $P$ and $Q$ have a leading coefficient of 1, but good spot that $P'$ may not, I have updated my question. – keith May 24 '16 at 12:05
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    If I am right, $C=1$. –  May 24 '16 at 12:07
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    @Yves Daoust, yes you are right, $C$ will indeed be 1 :-) – keith May 24 '16 at 12:09
  • Not sure that helps, but one can consider the roots of the pencil of polynomials $(1-\mu) P(x)+\mu Q(x)$, which evolves between the set of roots of $P$ (for $\mu=0$) and that of $Q$ ($\mu=1$), and corresponds to the remainder $P'$ when $\mu=\infty$. Maybe there is a direct way to interpolate between these root sets. –  May 24 '16 at 12:19
  • See this related question on MathOverflow http://mathoverflow.net/q/30072 –  May 24 '16 at 12:28
  • @Yves Daoust, yes I think you're right, it boils down to there being no general solution to the sum of two sets of roots and what the resulting set of roots looks like. Thanks for helping me think it through. – keith May 24 '16 at 12:38

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