0
$\begingroup$

For rational $f$ define the $n$th iterate

$$f_{n+1}(z) = f(f_n(z)), f_1(z) = f(z)$$

Show if $p$ is a polynomial of degree $\deg p > 1$, there is $R > 0$ so

$$|z| > R \implies p_n(z) \to \infty$$

Can anyone please let me know how to solve this? Thanks in advance.

$\endgroup$
  • $\begingroup$ I've transcribed the text, but I think there's a problem with the definition. Shouldn't it be $f_{n+1}(z) = f(f_n(z))$? $\endgroup$ – Alexis Olson Sep 30 '16 at 6:01
  • $\begingroup$ yeah it seems to be f n+1 $\endgroup$ – kkr Sep 30 '16 at 6:03
1
$\begingroup$

Let $p(z) = \sum_k a_k z^k$. Then $|p(z)| \geq |a_n z^n| + |\sum_{k=0}^{n-1} a_k z^k|$. We have some $R$ such that $|\sum_{k=0}^{n-1} a_k z^k| \leq \frac{|a_n|}{2}|z^n|$ for all $|z| > R$. We conclude for $|z| > R$ we have $|p(z)| \geq \frac{|a_n|}{2} |z|^n$. There is also a $\tilde{R}$ such that $|z| > \tilde{R}$ implies $\frac{|a_n|}{2} |z|^n > \tilde{R}$ (this is where we need $n \geq 2$). Now taking the maximum of $R$ and $\tilde{R}$ the claim follows.

$\endgroup$

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