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$P_5(w) = c_0 + c_1w + c_2w^2 +...+ c_5w^5$, where $ c_0, ... , c_5 $are 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} (Q_5(w))^2 + 16Q_5(w)(1 + w)^3 w^2 - 80(1 + w)^2 w^3 P_5(w) = 0 \end{equation} is also a root of \begin{equation} (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}

Is there a way to solve this using an algebra software?

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  • $\begingroup$ Every root (...) is also a root of (...) : with multiplicity? (In which case, you are stating that the latter polynomial is a multiple of the former). $\endgroup$ – Pietro Majer Sep 25 '12 at 20:37
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    $\begingroup$ Maybe a few words about the origin of this question would be nice. The question looks a little like the search for a degree $5$ cover of Riemann spheres with some ramification requirements. $\endgroup$ – Peter Mueller Sep 26 '12 at 13:41
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As @Pietro says, if the polynomial has no multiple roots, then the latter polynomial is a multiple of the former, which gives a system of polynomial equations satisfied by the coefficients. If the polynomial DOES have multiple roots, then you get some extra equations (for example, the discriminant is zero, and the polynomial factors, AND some of the factors are also factors of the second polynomial. In any event, however, the system you get is very high degree with many variables, so most likely out of reach.

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  • $\begingroup$ The idea is just to divide the second polynomial f_2(w) = (24)3(1+w)5w4[4P5(w)w−Q5(w)(1−w)]3+108(Q5(w))5 , to the tenth power by f_1 (w) = (Q5(w))2+16Q5(w)(1+w)3w2−80(1+w)2w3P5(w) then set the rest to zero. We should get up to nine conditions on c_i's. Using they are integers, I hope to get strong restrictions, or in the best case, one solution that is expected to be (1, 5 ,10 , 10 , 5, 1). $\endgroup$ – Renato Sep 26 '12 at 22:52
  • $\begingroup$ What do you mean by "set the rest to zero"? $\endgroup$ – Igor Rivin Sep 26 '12 at 23:36
  • $\begingroup$ Apparently Renato doesn't consider multiplicities. As $f_1(w)$ has degree at most $10$, his condition then is that the remainder of $f_2(w)^{10}$ modulo $f_1(w)$ vanishes. I believe `set the rest to zero' means the condition coming from setting the coefficients of this remainder to $0$. By the way, a solution to the question is $P(w)=w((w+1)^4+c)$ for any $c$. Also, $P(w)=-w(w+1)^4$ is a solution. Still, it would be nice to learn more about the background of this question! $\endgroup$ – Peter Mueller Sep 27 '12 at 9:59
  • $\begingroup$ Sorry, rest = reminder , as Peter said "the remainder of $f_2(w)^10 $ modulo $f_1(w)$ vanishes". The origin of this question comes from Mirror Symmetry and computing the superpotential $W$, which is just function, in my case from $C^2$ to $C$, 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 $W = z + \frac{2(1+w)^2}{z^2} + \frac{P_5(w)}{w z^5}$ – $\endgroup$ – Renato Sep 27 '12 at 22:21
  • $\begingroup$ From Floer theory we have that if $(z,w)∈critW$ then $W^3(z,w)=27$. Solving for z in one of the partial derivatives of W, we get that this condition is equivalent to the roots of $f_1(w)$ being also roots of $f_2(w)$. $\endgroup$ – Renato Sep 27 '12 at 22:41

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