# Power series expansion of the Koenigs function

Given a non-zero holomorphic function $f$ fixing $0$ which isn't a Mobius transform, the Koenigs function of $f$, which we'll call $h$, is the function which linearizes $f$ in the sense that $$h(f(z)) = f'(0)h(z).$$ I am interested in finding an expression or estimates for $h$ in the case where $f$ is a polynomial of low degree with real coefficients. (I'd be pretty happy with degree 3)

I have heard that there is a way of guessing the power series of $h$. Does anyone know of any references which explain how to do this?

• Can't you just write down a power series for $h$ and equate coefficients, starting $h(z)=z+h''(0)z^2/2+\ldots$?
– user25199
Feb 11 '14 at 18:19
• @Carl I'm not sure what you mean, what am I equating the coefficients of the power series of $h$ with? Feb 11 '14 at 18:46
• @Carl, I think I see what you mean now, I have been able to use it to get a recurrence relation for the coefficients. Thank you. Feb 12 '14 at 5:12

You do not tell the crucial thing: how large is $|f'(0)|$. There is no simple expression for coefficients or any other simple expression for $h$, even when $f$ is quadratic polynomial $\lambda z+z^2$.

However the global behaviour of $h$ has been be studied a lot, with remarkable results.

In the following description, I assume for simlicity that $f$ is a polynomial.

If $|f'(0)|<1$, $h$ is analytic in a neighborhood of $0$. It has an analytic continuation to the immediate domain of attraction of $0$, the boundary of this domain is the natural boundary for $h$, and the boundary behaviour of $h$ is relatively well understood.

If $|f'(0)|$ is a root of unity, the equation has no solution, even as a formal series.

If $|f'(0)|=1$ but not a root of unity, a formal power series $h$ exists. Let $f'(0)=e^{2\pi i\alpha}$, then everything depends on the fine Diophantine properties of $\alpha$. For quadratic polynomial $f$, there is a necessary and sufficient condition in terms of $\alpha$ for $h$ to be analytic in a neighborhood of $0$ (Yoccoz got a Fileds medal for this). When analytic, its maximal domain of analyticity is the so-called Siegel disc, and $h$ maps this Siegel disc onto a round disc bijectively.

When $|f'(0)|>1$, $h$ exists but its analytic continuation is not single valued. However the inverse of $h$ has a nice property: it is an entire function of finite order. This result is due to Poincare and Valiron.

For the introduction to the properties of this function see P. Fatou, Sur les equations fonctionnelles, in 3 parts, available on http://www.math.purdue.edu/~eremenko/books-papers.html and G. Valiron, Fonctions analytiques, Paris, Press Universitaires, 1954. For the case $|f'(0)|=1$, see the work of Yoccoz.

It follows from this brief description that there can be no simple expression for $h$ (whatever polynomial $f$ of degree at least $2$ you use). It is easy to obtain a recurrence for coefficients, but it is very complicated, and can hardly be used to study $h$. One exception is the work of Siegel on the case $|f'(0)|=1$. he was able to show that the radius of convergence of the power series is positive for certain $\alpha$ by a kind of direct analysis of this recurrence.

• Thanks for the answer, this is very interesting and very useful. For the family of polynomials I'm working with, $0<f'(0)<1$, so they are relatively well behaved. Feb 12 '14 at 5:32