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The practical question is: $P_5(w) = c_0 + c_1w + c_2w^2 +...+ c_5w^5$, where $ c_0, ... , c_5 $ are positive integers that I want to determine.

$Q_5(w) = w^2 (\frac{d}{dw}( \frac{P_5(w)}{w}) = -c_0 + c_2w^2 + 2c_3w^3 + 3c_4w^4 +4c_5w^5$, another deg 5 polynomial.

I have that every root (over $\mathbb{C}$) of \begin{equation} f_1(w) = (Q_5(w))^2 + 16Q_5(w)(1 + w)^3 w^2 - 80(1 + w)^2 w^3 P_5(w) = 0 \end{equation} that is not a root of Q_5(w) neither w = 0 or w =-1, is also a root of \begin{equation} f_2(w) = (24)^3 (1 + w)^5 w^4 [4P_5 (w) w - Q_5(w) ( 1 + w)]^3 + 108(Q_5(w))^5 = 0 \end{equation}

So I want to divide f_2(w)^{10} by f_1(w) and set the remainder to zero to get conditions over the c_i's. I don't know the best algebra software to use neither how to implement the algorithm.

Context: The origin of this question comes from Mirror Symmetry and computing the superpotential $W$, which is just function, in my case from $\mathbb{C}^2$ to $\mathbb{C}$, of the Landau-Ginzburg model for the complement of a divisor in $CP^2$. The Superpotential is a Laurent polynomial whose coefficients are given by a count of holomorphic discs, and monomials are related to a relative homotopy class of the group. I was able to prove that $$W= z+\frac{2(1+w)^2}{z^2}+\frac{P_5(w)}{w z^5}$$ Using Floer theory one can show that the critical points (z,w) of $W$ (which are in correspondence with roots of $f_1(w)$ that are not root of $Q_5(w), w $ and $w+1$) satisfies W(z,w) = 27, from where we get f_2(w) = 0.

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  • $\begingroup$ Typo: on "relative homotopy class of the group" read relative homotopy class of the disc. $\endgroup$ – Renato Sep 27 '12 at 23:21
  • $\begingroup$ For "So I want to divide $f_2(w)^{10}$ by $f_1(w)$ and set the remainder to zero to get conditions over the c_i's. I don't know the best algebra software to use neither how to implement the algorithm. I am using the assumption that every root of $f_1(w)$ is a root of $f_2(w)$. $\endgroup$ – Renato Sep 27 '12 at 23:31
  • $\begingroup$ en.wikipedia.org/wiki/Berlekamp%E2%80%93Massey_algorithm $\endgroup$ – Steve Huntsman Sep 28 '12 at 5:36

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