Suppose $f(z) = P(z)e^{Q(z)}$ where $P,Q$ are real polynomials. What is the number of non-real zeros of $f^{(k)}$ as $k$ increases?
We know that $f''$ has $\geq m$ zeros where $m$ depends on $Q(z)$.
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4$\begingroup$ This could be interesting -- it might be worth adding some motivation, as well as your state of knowledge about the problem, however. $\endgroup$– Daniel LittCommented Jul 26, 2010 at 17:52
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$\begingroup$ I can't edit - someone should put a $k$ between "as" and "increases". $\endgroup$– Gerry MyersonCommented Jul 26, 2010 at 22:13
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2 Answers
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Just some hints. The functions $f^{\,(k)}$ have the same zeros of the polynomials $P_k:=f^{\,(k)}\exp(-Q)$, that satisfy $P_0:=P$ and $P_{k+1}=P_k'+P_kQ'$. In particular $P_k$ has degree $\deg(P)+k\left(\deg(Q)-1\right)$, and this is also the total number of zeros of $f^{\,(k)}$. They may be all real: for instance if $Q:=-x^2$ and $P:=1$ one finds the Hermite polynomials, that are orthogonal, hence have all zeros real and simple.
answered Jul 26, 2010 at 22:37
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One could use the level sets $\{z \in H^{+}: \text{Im} Q(z) = 0 \}$ to count non-real zeros of the derivatives of $f(z)$ where $Q$ is the Newton's method function for $f$.
