Suppose $f$ is a holomorphic function in a neighborhood of zero fixing zero. Suppose $f'(0) = \lambda$ and $0<\lambda < 1$. It's not so hard to prove that $f^{\circ k}(z) = f(f(\ldots\text{$k$ times}\ldots f(z))) \sim \lambda^k \Psi(z)$ as $k\to\infty$, where $\Psi(z)$ is the Schröder function of $f$ satisfying $\Psi(f(z)) = \lambda \Psi(z)$ and $\Psi'(0) = 1$. (See for instance John Milnor's "Dynamics in One Complex Variable")

Recently I've encountered a kind of binomial expansion. Let

$$I_n(z) = \sum_{k=0}^n \binom{n}{k}(-1)^kf^{\circ k}(z)$$

It seems intuitive that since $f^{\circ k}$ looks like $\lambda^k$, $I_n$ should look like $(1-\lambda)^n$. Additionally $I'_n(0) = (1-\lambda)^n$, so the heuristic plays fairly well. Sadly I'm having trouble proving this.

With that being said, my question can be asked:

Is $I_n(z) \sim (1-\lambda)^n \Psi(z)$ as $n\to \infty$?

If this proves too strong a statement, I'll settle for the more relaxed statement:

$$|I_n(z)| < Cr^n$$

for some $0<r<1$ and an arbitrary constant $C$.

If both of these prove too strong,

What can we say about the asymptotics of $I_n$?

Any help would be greatly appreciated.

Thanks, Richard.


1 Answer 1


Is this too easy? We have $f^k(z) = \Psi^{-1}(\lambda^k\Psi(z))$, where $\Psi^{-1}(z)=z+\sum_{i=2}^\infty a_iz^i$. Substituting yields $$ \sum_{k=0}^n\binom{n}{k}(-1)^kf^k(z) = (1-\lambda)^n\Psi(z) + \sum_{k=0}^n\binom{n}{k}(-1)^k\sum_{i=2}^\infty a_i(\lambda^k\Psi(z))^i. $$ The error term is $$ \sum_{k=0}^n\binom{n}{k}(-1)^k\sum_{i=2}^\infty a_i(\lambda^k\Psi(z))^i = \sum_{i=2}^\infty a_i \Psi(z)^i \sum_{k=0}^n\binom{n}{k}(-1)^k \lambda^{ik} = \sum_{i=2}^\infty a_i \Psi(z)^i (1-\lambda^i)^n. $$ So if you fix a (small) $z$, divide by $(1-\lambda)^n$, and let $n\to\infty$, your error term looks like $$ \sum_{i=1}^\infty a_i\Psi(z)^i(1+\lambda+\lambda^2+\cdots+\lambda^{i-1})^n. $$ It's hard to how this is ever going to be $\ll(1-\epsilon)^n$ as $n\to\infty$, if you insist that $\epsilon$ be independent of $z$. If you also let $z\to0$ appropriately as $n\to\infty$, then since $\Psi(z)\to0$, you can make it work.

  • $\begingroup$ I think you're on to something here. The more I mess around with this the more I'm doubting its truth. I think I need further conditions on $f$ to get this to work; or I need to finesse the asymptotics more. Frankly, I could only prove $\mathcal{I}_n$ is bounded as $n \to \infty$. I can't even prove it converges to $0$. I may have to do a work around, and take a more complicated path. Thanks for your answer though, I don't know why I didn't think of using the Schroder function more explicitly. I was approaching this using the Mellin transform, and the identity $I_{n+1}(z) - I_n(z) = I_n(f(z))$ $\endgroup$ Feb 7, 2019 at 3:34

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