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I'm sorry if the question is too vague but maybe someone can help me. I am working with a polynomial $f(x)$ with distinct roots $x_1,\ldots,x_d$. I am able to solve the problem I am currently working on if it turns out that $f''(x_i)=0$ for a certain $i\in{1,\ldots,d}$, i.e. $f(x)$ shares a root with its second derivative.

Then I'd like to know what I can say about $f(x)$ knowing that $f'(x_i)$, $f''(x_i)\neq 0$ $\forall i=1,\ldots,d$.

Any suggestion or deduction would be helpful since I fear I am overlooking something or perphaps this condition is too weak to be useful.

EDIT: I already have some algebraic relations among the roots (any cross-ratio lies in a finite set) and I'd like to obtain informations about the coefficients of $f$; of course if I could find an algebraic relation for every single root it would be over.

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You can check if two polynomials share a root by taking their resultant. –  John Wiltshire-Gordon Nov 4 '12 at 22:11
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What do you WANT to say about $f$? –  Alexandre Eremenko Nov 5 '12 at 0:48
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1 Answer

The resultant $R$ of two polynomials (a polynomial in the coefficients of the two polynomials) vanishes iff the two polynomials share a common root. Thus $R(f,f'')=0$ and $R(f,f')\ne 0$ are your conditions. Thus the polynomials that you want form a hyper surface in a Zariski open domain.

Added:

Looking at $R(f,f'')$ one sees that as a polynomial in one coefficient $f_i$ it is of order $d-1$ in $f_i$ (keeping the other coefficients fixed). So, if you vary only the constant term $f_0$, say, you find $d-1$ intersection points of this line with $R(f,f'')=0$, disregarding the requirement that $R(f,f')\ne 0$. This you regain by wiggling the line to a (complex) curve which avoids the hyper surface $R(f,f')=0$.

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