Is $\lim_{n \to \infty} f^{[n]} ( g^{[n]} (z) ) $ analytic?

Question

Let $^{[*]} $ denote composition. Let $f(z) = \sqrt z$ and $g(z) = z^2 + 1$ Define $Q(z)$ as

$$ Q(z) = \lim_{n \to \infty} f^{[n]} ( g^{[n]} (z) ) $$

Is $\lim_{n \to \infty} f^{[n]} ( g^{[n]} (z) ) $ analytic for $Re(z) > 11$ ?

Answer 1

Yes, $$Q(z) = \lim_{n \to \infty} f^{[n]} ( g^{[n]} (z)); \;\;\;g(z)=z^2+1; \;\;\;f(z)=\sqrt{z}$$

is analytic and well defined so long as $|g^{[n]}(z)|$ gets arbitrariliy large. When I have more time, I'll calculate the Laurent series for the limit of the composition and add it to this post. see below

Most points grow arbitrarily large where the limit is defined. But there should be a set of Cantor dust points on the Julia set boundary for which the composition is not defined. For example, $z=\pm 1/2 \pm i\sqrt{7/4}\;\;z=\pm 1/2 \pm i \sqrt{3/4}$ are eight of the points in the Cantor dust where the limit is not defined, since at these points, iterating g goes to a fixed point or a two-cycle. I'll provide more details when I have more time.

continued What you are looking for is the Bottcher function for g(z), where we move the fixed point from infinity to zero. So, instead of iterating $z \mapsto g(z)$, we iterate $z \mapsto \frac{1}{g(1/z)}$

Call this new function $$g2(x)=\frac{1}{x^{-2}+1} = x^2-x^4+x^6-x^8+x^{10}-x^{12}...$$ Then g2 has super-attracting fixed point at x=0, and therefore has a formal Bottcher series, which turn's out to be the Op's desired function, whose formal series solution is posted below where $f(x)=\sqrt{x}\;$ since $\sqrt{x}=\frac{1}{\sqrt{1/x}}$

$$B(x) = \lim_{n \to \infty} f^{[n]} ( g2^{[n]} (x) )$$ Notice that B(x) meets the definition of the Bottcher function, $B(g2(x)) = B(x)^2$, where we are putting iterating g2(x) into correspondence with iterating x^2. Finally, B(x) leads to a Laurent series for the Op's desired analytic function.

$$Q(x) = \frac{1}{B(1/x)} = \lim_{n \to \infty} f^{[n]} ( g^{[n]} (x) ) $$ As noted, the function is not defined at an uncountable set of points corresponding to a Cantor dust. See Complex Dynamical Systems edited by Devaney, chapter 1, "The Complex dynamics of quadratic polynomials"

{B=
 x
+x^ 3* -1/2
+x^ 5*  1/8
+x^ 7*  5/16
+x^ 9* -101/128
+x^11*  321/256
+x^13* -1515/1024
+x^15*  1677/2048
+x^17*  57235/32768
+x^19* -440827/65536
+x^21*  3216383/262144
+x^23* -6517069/524288
+x^25* -7260785/4194304
+x^27*  292133525/8388608
+x^29* -2439251715/33554432 ...}

I noticed the Op also asked if $Q(z)$ defined/analytic $\forall \; \Re(z)>11$? I think 11 was probably a typo and the Op meant $\forall \; Re(z)>1?$

The answer is yes, since it is not too difficult to show iterating $$g(z)=z^2+1;\; \forall\; \Re(z)\ge 1;\; |g(z)| \ge \sqrt{4+\Im(z)^4} \ge 2$$

And once we have $|g(z)| \ge 2$ we can trivially show iterating $g(z)$ one more time leads to $|g(g(z))|\ge 3$, and further iterations get arbitrarily large where the Botcher function is always defined. So then all of the points in the Cantor dust have $|\Re(z)|<1\; \text{and}\; |\Im(z)|<2$.