2
$\begingroup$

Edit: My serious thanks to Christian, Todd and Yemon for their comments to the previous version. The derivative is the same as Todd says: $(az^{n})'=naz^{n-1}$. The polynomial is in the form of $\sum a_{i}z^{i}$ not in the form of $\sum z^{i}a_{i}$, but the comment of Christian is a very interesting comment

Motivated by an equivalent formulation of the Gauss Lucas theorem in terms of half planes, we ask:

Assume that all roots of a polynomial $p(z)$ with Quaternions coefficients is contained in the half space $\{(x_{1},x_{2},x_{3},x_{4})\mid x_{4}\geq0 \}$.Is the same true for $p'(z)$?

Improve this question
$\endgroup$
5
  • 4
    $\begingroup$ These are non-commutative polynomials? And, by the way, what is the derivative? $\endgroup$
    – Gerald Edgar
    Commented Jun 19, 2016 at 23:36
  • 2
    $\begingroup$ I assume the formal derivative where $(z^n)' = n z^{n-1}$ is meant. But it is worth remarking that noncommutative polynomials behave very differently, e.g., non-unique factorization. As for example how $z^2 + 1$ has continuum many roots. $\endgroup$
    – Todd Trimble
    Commented Jun 19, 2016 at 23:53
  • 2
    $\begingroup$ Another issue is where to put the coefficients. Or put differently, is it clear that $\sum a_j z^j$ gives the same functions as $\sum z^j b_j$ ? $\endgroup$
    – Christian Remling
    Commented Jun 20, 2016 at 0:10
  • 1
    $\begingroup$ Downvoting for now, until Ali clarifies what he means by a quaternionic polynomial and its derivative $\endgroup$
    – Yemon Choi
    Commented Jun 20, 2016 at 0:24
  • $\begingroup$ If we start with a polynomial with real coefficients, then we avoid the issue with the non-commutativity of $\sum a_j z^j $ vs. $\sum z^j a_j$, and the derivative makes sense. So perhaps there is something interesting to say here? (I think these zero sets might not lie in half-spaces of the form $x_i\geq 0$, but you could ask for arbitrary half-spaces.) $\endgroup$
    – Harry Richman
    Commented Feb 23, 2017 at 16:56

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .