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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.

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.

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 $f^{\circ k}(z) = f(f(...k\,times...f(z))) \sim \lambda^k \Psi(z)$ 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.

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.

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 $f^{\circ k}(z) = f(f(...k\,times...f(z))) \sim \lambda^k \Psi(z)$ 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 $(1-\lambda)^n$. 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.

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 $f^{\circ k}(z) = f(f(...k\,times...f(z))) \sim \lambda^k \Psi(z)$ 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.