Conjugation near a super-attractive fixed point

Section 5.3 Conjugation near a super-attractive fixed point

We now address the super attracting case. While there are some superficial similarities in the dynamics between the an attracting fixed point and a super-attracting fixed points, there are also substantial differences - starting with the analytical description of \(f\text{.}\) If \(z_0\) is a super-attracting fixed point of \(f\text{,}\) then (expanding \(f\) in a Taylor series about \(z_0\)) we have

\begin{equation*} f(z) = z_0 + a_m(z-z_0)^m + a_{m+1}(z-z_0)^{m+1}+\cdots \end{equation*}

for some integer \(m\geq 2\text{.}\) It's immediately apparent that \(f\) is an \(m\)-to-1 function for every small neighborhood of \(z_0\text{.}\)

It will simplify matters a little bit to suppose that the fixed point is zero and that \(a_m=1\text{.}\) The fact that we can do so without loss of generality is established in exercises Exercise 5.6.1 and Exercise 5.6.2. At the simplest level, a super-attractive fixed point is contained in a neighborhood where all the points converge to that fixed point under iteration; this is analogous to lemma Lemma 5.1.1

Lemma 5.3.1

Suppose that \(f\) can be written as

\begin{equation*} f(z) = z^m + a_{m+1}z^{m+1}+\cdots \end{equation*}

in a neighborhood of the origin, so that zero is a super-attracting fixed point of order \(m\) for \(f\text{.}\) Then, there is a neighborhood of the origin where all the points of converge to zero under iteration of \(f\text{.}\)

Proof

By Taylor's theorem, there are positive constants \(C\) and \(r\) such that

\begin{equation*} |f(z)| < C|z|^m \end{equation*}

for all \(z\) such that \(|z| < r\text{.}\) By induction, for such \(z\)

\begin{equation*} |f^n(z)| < C|z|^{m^n}. \end{equation*}

The result follows.

Theorem 5.3.2

Suppose that \(f\) can be written as

\begin{equation*} f(z) = z^m + a_{m+1}z^{m+1}+\cdots \end{equation*}

in a neighborhood of the origin, so that zero is a super-attracting fixed point of order \(m\) for \(f\text{.}\) Then, \(f\) is analytically conjugate to \(g(z)=z^m\text{.}\)

Like theorem Theorem 5.1.2, this theorem can be appreciated on multiple levels. For the time being, we satisfy ourselves with the construction.

Proof

Let

\begin{equation*} \varphi_n(z) = f^n(z)^{m^{-n}}, \end{equation*}

which, by the inequalities in the proof of lemma Lemma 5.3.1, is well defined in some neighborhood of the origin. Note that

\begin{equation*} \varphi_{n-1}\circ f = (f^{n-1}\circ f)^{m^{-(n-1)}} = g\circ\varphi_n. \end{equation*}

Thus, if \(\varphi_n\) converges to \(\varphi\text{,}\) then \(\varphi\circ f = g\circ\varphi\text{.}\)

Figure Figure 5.3.3 illustrates a this conjugacy for \(f(z)=z^2-z^4/8\text{.}\) Note that the origin is a super-attracting fixed point - but not the only one. In fact, this map was created expanding \((z^2-1)^2-1\) and then conjugating to force the coefficient of \(z^2\) to be one. Note also that the conjugacy extends only through the immediate basin of attraction of the origin. The curves that we see in side that basin are the inverse images of rays of constant argument.