arXiv subjects

Vasiliki Evdoridou

Publications and source records attributed to Vasiliki Evdoridou.

21 records · Page 2Linked to original sources

arXiv2016

Regularity and growth conditions for fast escaping points of entire functions

Let $f$ be a transcendental entire function. The quite fast escaping set, $Q(f)$, and the set $Q_2(f),$ which was defined recently, are equal to the fast escaping set, $A(f),$ under certain conditions. In this paper we generalise these sets by introducing a family of sets $Q_m(f)$, $m \in \mathbb{N}.$ We also give one regularity and one growth condition which imply that $Q_m(f)$ is equal to $A(f)$ and we show that all functions of finite order and positive lower order satisfy $Q_m(f)=A(f)$ for any $m$. Finally, we relate the new regularity condition to a sufficient condition for $Q_2(f)=A(f)$ introduced in recent work.

math.DS↗

arXiv2015

Fatou's web

Let $f$ be Fatou's function, that is, $f(z)= z+1+e^{-z}$. We prove that the escaping set of $f$ has the structure of a `spider's web' and we show that this result implies that the non-escaping endpoints of the Julia set of $f$ together with infinity form a totally disconnected set. We also give a well-known transcendental entire function, due to Bergweiler, for which the escaping set is a spider's web and we point out that the same property holds for families of functions.

math.DS↗

arXiv2015

Fast escaping points of entire functions: a new regularity condition

Let $f$ be a transcendental entire function. The fast escaping set, $A(f)$, plays a key role in transcendental dynamics. The quite fast escaping set, $Q(f)$, defined by an apparently weaker condition is equal to $A(f)$ under certain conditions. Here we introduce $Q_2(f)$ defined by what appears to be an even weaker condition. Using a new regularity condition we show that functions of finite order and positive lower order satisfy $Q_2(f)=A(f)$. We also show that the finite composition of such functions satisfies $Q_2(f)=A(f)$. Finally, we construct a function for which $Q_2(f) \neq Q(f)= A(f)$.

math.DS↗