1
$\begingroup$

The (probably) famous Gauss-Lucas theorem states that the roots of the derivative $P'(z)$ are contained in the convex hull of the roots of $P(z)$, where $P(z)$ is complex variable polynomial.

I am interested here in could it be the case that we always have some polynomial of any degree (except $1$) $P(z)$ such that some root of its derivative is "at a small as we want distance" from some root of $P(z)$.

To be more precise, here is the statement of the question:

Is it true that for every $\varepsilon>0$ and for every $n\in \mathbb N \setminus \{1\}$ there exists polynomial $P(z)$ in complex variable of degree $n$ with $n$ different roots such that there is root $z_a$ of $P'(z)$ and root $z_b$ of $P(z)$ which are such that we have $|P'(z_a)-P(z_b)|<\varepsilon$

|cite|improve this question
$\endgroup$
  • $\begingroup$ Yes, start with a polynomial with a double root and perturb the coefficients. $\endgroup$ – Michael Burr Nov 14 '15 at 1:11
  • $\begingroup$ @MichaelBurr All roots are different in the question, if that changes anything? $\endgroup$ – Farewell Nov 14 '15 at 1:13
  • $\begingroup$ If you insist that the coefficients of the polynomial are bounded integers, then the answer is no. $\endgroup$ – Michael Burr Nov 14 '15 at 1:13
  • $\begingroup$ No, it doesn't. After perturbation, the roots will all be different, but because the roots depend continuously on the coefficients, there will be a root of the derivative arbitrarily close to a root of the function. $\endgroup$ – Michael Burr Nov 14 '15 at 1:14
  • $\begingroup$ @MichaelBurr If we choose some $\varepsilon_0>0$ is there an easy way to construct some concrete polynomial for every degree $n>1$ (it is a sequence of polynomials for every concrete $\varepsilon>0$ ) such that the question holds? I do not see immediately that this is the case. $\endgroup$ – Farewell Nov 14 '15 at 1:20
7
$\begingroup$

Let $P(z) = (z-\epsilon)(z-2\epsilon)\cdots (z-n\epsilon).$ Using the mean value theorem we can see that every root of $P'$ is less than $\epsilon$ away from some root of $P.$

|cite|improve this answer
$\endgroup$
2
$\begingroup$

Let $P(z)=(z-\epsilon)(z+\epsilon)\prod_{i=1}^{n-2}(z-i)$. This is a perturbation of the polynomial $z^2\prod_{i=1}^{n-2}(z-i)$, removing the double root.

|cite|improve this answer
$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.