Decay of the binomial expansion of $f^{\circ k}$

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 $$0 and an arbitrary constant $$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.

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.
• 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 $I_{n+1}(z) - I_n(z) = I_n(f(z))$ Feb 7, 2019 at 3:34