1
$\begingroup$

$\newcommand{\CC}{\mathbb{C}}$

Let for $n > 1$ and $m = n-1$ $$ p = x^n + a_1 x^{n-1} + \cdots + a_m x $$ be a polynomial with $a_i \in \CC$. Call $p^{(i)}(x) = \frac{d^ip}{dx^i}(x)$.

The following question would have an impact on the problem, if sequences of resultants $\mathrm{res}_x(p(x), p^{(n-i)}(x))$, $i = 1,\ldots,s$ are regular sequences:

Take $s < m$ and consider the space $X$ of monic polynomials $p$ as above such that $\alpha_1,\ldots,\alpha_s \in \CC$ (not necessarily distinct, could be even all equal) exist so that

\begin{equation} p^{(n-1)}(\alpha_1) = p^{(n-2)}(\alpha_2) = \cdots = p^{(n-s)}(\alpha_s) = 0 \end{equation}

and \begin{equation} p(\alpha_1) = \cdots = p(\alpha_s) = 0 \end{equation}

Question:

Can one show that for a certain solution $p(x) \in X$ with $a_{s+1} = 0$ there exists in every neighbourhood (identifying $X \cong \CC^m$) a solution in $X$ with $a_{s+1} \neq 0$?

This is obviously the case for $2 s + 1 \leqslant m$ where one can think of the $\alpha_i$ as fixed and consider the problem as a statement of linear algebra. If $2 s + 1 > m$ it may become necessary to "move the $\alpha_i$ a bit", to get $a_{s+1} \neq 0$.

I did some computer calculations with Maple and it always seemed to be possible to do such a small move, leading to $a_{s+1} \neq 0$, but I could not find a general proof.

Is the answer to the above question maybe already known to specialists in (Birkhoff-)interpolation? What could be helpful to read for making progress on this problem?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.