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How can we prove the following?

If $\frac{dP_{n}}{dz}|_{z=z_{0}}=0$ then $|P_{n}(z_{0})|<2$ for all $n>1$, where $P_{n}(z)\equiv P_{n-1}^{2}+z$ and $P_{1}\equiv z$

$z$ is in the complex plane.

It appears that $\lim{}_{n\rightarrow\infty}\max\{|P_{n}(z_{0})|:P_{n}'(z_{0})=0\}=2 $, but all I really need is a proof that it's always <2 not that it approaches 2.

Both are easy to demonstrate with the Mathematica code below, but I can't figure out a proof.

p[n_, z_] := If[n > 1, p[n - 1, z]^2 + z, z];
Do[Print[Max[Abs[p[n, z] /. Solve[D[p[n, z], z] == 0, z]] // N]], {n, 2, 8}]

Output:

0.25

1.15268

1.76235

1.94118

1.98545

1.99638

1.9991

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  • $\begingroup$ I apologize, but there is something that I do not understand: the recurrence gives $P_1 = z, P_2 = z^2 + z, P_3 = z^4 + 2z^3 + z^2 + z, \dots$, so the corresponding derivatives in $z_0$ are $1, 2 z_0 + 1, 4 z_0 ^3 + 6 z_0 ^2 + 2 z_0 + 1, \dots$; how can these all be $0$ in $z_0$? $\endgroup$
    – Alex M.
    Nov 13, 2015 at 5:38
  • $\begingroup$ @AlexM. Each $P_n$ will have a different set of $z_0$'s. Note in the Mathematica code that the $z_0$'s are recalculated for each value of n. $\endgroup$ Nov 13, 2015 at 6:22
  • $\begingroup$ I must ask REALLY hard questions, because this is the third time I've posted a bounty no one won. $\endgroup$ Nov 20, 2015 at 3:16
  • $\begingroup$ Some observations. All real $z_0$ are in $(-2,0)$. So to prove the inequality in reals, it suffices to show that $|P_n[z]| \le 2$ for $z \in (-2,0)$, which should be true by numeric evidence. Another observation is that on this range $P_n(z) \le P_2(z)$. $\endgroup$
    – ablmf
    Nov 27, 2019 at 15:36
  • $\begingroup$ @ablmf Well, actually, $P_n$(-2<=z<=1/4) <= 2 for all n and real z. But I need to prove this for complex z. $\endgroup$ Nov 27, 2019 at 22:19

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