# Dynamics

Now we focus on the complex plane and it's dynamics. The composition $$f(g(z))$$ can construct any iterated function at a fixed point, not at infinity, by setting $$f(0)=0$$ and $$g(z)=f^{t-1}(z)$$ with $$t \in \mathbb{N^+}$$. The final proof of this section extends $$t \in \mathbb{C}$$. The first derivative of $$f(z)$$ at its fixed point $$f'(0)$$ is often represented by $$\lambda$$ and referred to as the multiplier or the Lyapunov characteristic number; its logarithm is known as the Lyapunov exponent.

### The First Derivative

The first derivative results in the well known Linearization Theorem.

\begin{align} & Df(g(z)) = f'(g(z))g'(z) \nonumber\\ & =f'(f^{t-1}(z))Df^{t-1}(z) \nonumber\\ & =\prod^{t-1}_{k_1=0}f'(f^{t-k_1-1}(z))\nonumber \end{align}

$$Df^t(0)=\lambda^t$$

### The Second Derivative

The higher derivatives can be evaluated in an analogous fashion from the second derivative.

\begin{align} D^2f(g(z)) & = f''(g(z))g'(z)^2+f'(g(z))g''(z) \nonumber\\ & = f''(f^{t-1}(z))(Df^{t-1}(z))^2+f'(f^{t-1}(z))D^2f^{t-1}(z)\nonumber \end{align}

Setting $$g(z) = f^{t-1}(z)$$ results in

$D^2f^t(0) = f''(0) \lambda^{2t-2}+\lambda D^2f^{t-1}(0)$

When $$\lambda \neq 0$$, a recurrence equation is formed that is solved as a summation.

\begin{align} D^2f^t(0) & = f''(0)\lambda^{2t-2}+\lambda D^2f^{t-1}(0) \nonumber\\ & = \lambda^0f''(0) \lambda^{2t-2} \nonumber\\ & + \lambda^1f''(0) \lambda^{2t-4} \nonumber\\ & + \cdots \nonumber\\ & + \lambda^{t-2}f''(0) \lambda^2 \nonumber\\ & + \lambda^{t-1}f''(0) \lambda^0 \nonumber\\ & = f''(0)\sum_{k_1=0}^{t-1}\lambda^{2t-k_1-2}\nonumber \end{align}

### The Third Derivative

Continuing on with the third derivative,

\begin{align} D^3f(g(z))& = f'''(g(z))g'(z)^3+3f''(g(z))g'(z)g''(z)+f'(g(z))g'''(z)\nonumber\\ & = f'''(f^{t-1}(z))(Df^{t-1}(z))^3\nonumber\\ & \quad +3f''(f^{t-1}(z))Df^{t-1}(z)D^2f^{t-1}(z)\nonumber\\ & \quad +f'(f^{t-1}(z))D^3f^{t-1}(z)\nonumber \end{align}

\begin{align} D^3f^t(0)&= f'''(0)\lambda^{3t-3}+3 f''(0)^2\sum_{k_1=0}^{t-1}\lambda^{3t-k_1-5} +\lambda D^3f^{t-1}(0) \nonumber\\ &= f'''(0)\sum_{k_1=0}^{t-1}\lambda^{3t-2k_1-3} +3f''(0)^2 \sum_{k_1=0}^{t-1} \sum_{k_2=0}^{t-k_1-2} \lambda^{3t-2k_1-k_2-5} \nonumber \end{align}

Note that the index $$k_1$$ from the second derivative is renamed $$k_2$$ in the final summation of the third derivative. A certain amount of renumbering is unavoidable in order to use a simple index scheme.

### The Fourth Derivative

Now the fourth derivative,

\begin{align} D^4f(g(z))&= f^{(4)}(g(z))g'(z)^4 +3f''(g(z))g''(z)^2 +f'(g(z))g^{(4)}(z) \nonumber\\ & \quad + 4g'(z) f''(g(z))g'''(z) + 6g'(z)^2 f'''(g(z))g''(z) \nonumber\\ &= f^{(4)}(f^{t-1}(z))(Df^{t-1}(z))^4 \nonumber\\ & \quad +3f''(f^{t-1}(z))(D^2 f^{t-1}(z))^2 \nonumber\\ & \quad +4Df^{t-1}(z) f''(f^{t-1}(z))(D^3 f^{t-1}(z)) \nonumber\\ & \quad +6(Df^{t-1}(z))^2 f'''(f^{t-1}(z))(D^2f^{t-1}(z)) \nonumber\\ & \quad +f'(f^{t-1}(z))D^{4} f^{t-1}(z) \nonumber \end{align}

\begin{align} D^4f^t(0)&= 12 f''(0)^3 \sum_{{k_1} = 0}^{t-1} \sum_{{k_2} = 0}^{t - {k_1}-2} \sum_{{k_3} = 0}^{t - {k_1} - {k_2}-3} {\lambda}^{4\,t - 3\,{k_1} - 2\,{k_2}- {k_3} -9} \nonumber\\ & \quad + 4 {f''(0)} {f'''(0)} \sum_{{k_1} = 0}^{t-1} \sum_{{k_2} = 0}^{t - {k_1}-2} {\lambda}^{4\,t - 3\,{k_1} - 2\,{k_2} -7}\nonumber\\ & \quad + 3 f''(0)^3 \sum_{{k_1} = 0}^{t-1} \sum_{{k_2}= 0}^{t - {k_1} -2} \sum_{{k_3} = 0}^{t - {k_1} -2} {\lambda}^{4\,t - 3\,{k_1} - {k_2} - {k_3} -8} \nonumber\\ & \quad + 6 {f''(0)} {f'''(0)} \sum_{{k_1} = 0}^{t-1} \sum_{{k_2} = 0}^{t - {k_1} -2} {\lambda}^{4\,t - 3\,{k_1} - {k_2} -6} \nonumber\\ & \quad + {f^{(4)}(0)} \sum_{{k_1} = 0}^{t-1} {\lambda}^{4\,t - {k_1} -4} \nonumber \end{align}

### The Higher Derivatives

Expanding out Faà di Bruno's equation gives,

$$D^n f^t(z)= \sum_{\pi(n)} \frac{n!}{k_1! \cdots k_n!} (D^k f)(f^{t-1}(z)) \left(\frac{Df^{t-1}(z)}{1!}\right)^{k_1} \cdots \left(\frac{D^n f^{t-1}(z)}{n!}\right)^{k_n}$$

The Taylor series of $$f^t(z)$$ is derived by evaluating the derivatives of the iterated function at a fixed point $$f^t(0)$$ by setting $$z=0$$ and separating out the $$k_n$$ term of the summation that is dependent on $$D^n f^{t-1}(0)$$.

The remaining $$\pi(n)-1$$ terms of the summation are only depend on $$D^k f^{t-1}(0)$$, where $$0<k<n$$.

Let this partial summation be written as $$\sigma(n)$$ with $$\sigma(0)=0$$ and $$\sigma(1) = 0$$.

Rewriting the $$\pi(n)-1$$ terms of the summation as $$\sigma(n)$$ will help in writing a proof by general induction. For $$n>1$$,

$D^n f^t(0)=\sigma(n) + \lambda D^n f^{t-1}(0) \blacksquare$

Theorem: Iterated Entire Function Theorem

The Taylor series of an iterated entire function $$f^t(z)$$ can be constructed given a fixed point and $$t \in \mathbb{N^+}$$.

Proof

Assume the function $$f(z)$$ is an entire function. Assume a fixed point at zero. As an entire function under composition, the Taylor series of $$f^t(z)$$ can be constructed for radius $$R$$ where $$0 < |z| < \infty$$ if and only if $$D^n f^t(0)$$ can be constructed for every $$n \geq 0$$.

Prove by strong induction.

Basis Steps:
Case $$n=0$$.
By definition $$D^0 f^t(0) = 0$$, so $$D^0 f^t(0)$$ can be constructed.
Case $$n=1$$.
Let $$D^1 f^t(0) = \lambda^t$$, so $$D^1 f^t(0)$$ can be constructed.
Case $$n=k-1$$.
Assume that $$D^k f^t(0)$$ can be constructed for all $$k$$ where $$0 \leq k < n$$. (Induction Hypothesis)
Induction Step:$$n=k$$.
Using the Dynamical Recurrance Equation, $$D^k f^t(0)=\sigma(k) + D'f(0) D^k f^{t-1}(0)$$. The function $$\sigma(k)$$ in only dependent on $$D^0 f(0), \ldots, D^k f(0)$$, and $$D^k f^t(0), \ldots, D^{(k-1)} f^t(0)$$. By the strong induction hypothesis, $$\sigma(k)$$ can be constructed. Therefore Eq. [1] can be reduced to a geometrical progression based on $$D'f(0)$$ that can be represented by a summation.

$D^k f^t(0) = \sum_{j=0}^{k-1} \sigma(k) \lambda^j$

This completes the induction step that $$D^n f^t(0)$$ can be constructed for all whole numbers $$n$$. $$\blacksquare$$

The Taylor series for $$f^t(z)$$ is

$f^t(z) = z_0+ \sum_{n=1}^\infty \sum_{j=0}^{n-1} \sigma(n) \lambda^j z^n$ Dynamical Equation

Theorem: Extension Theorem Given $$\lambda \neq 0$$ and $$\lambda^k \neq 1$$, the time variable $$t$$ in $$f^t(z)$$ can be extended from $$t \in \mathbb{N^+}$$ to $$t \in \mathbb{C}$$.

Proof

Let $$k \in \mathbb{N^+}$$.

Case $$|\lambda| \notin \{0,1\}$$: Schroeder's Functional Equation - Simplifies as a geometric progression.

Case $$\lambda = 1$$: Abel's Functional Equation - Simplifies as an arithmetic progression. $$\blacksquare$$.

### Convergence

Convergence can be proven by noting that there can be no finite points that are closest to the origin in $$t, z$$ where convergence doesn't hold.

Theorem: The iterated entire function $$f^t(z)$$ is convergent at all points in the complex plane except for infinity.

Assume $$\tau \in \mathbb{C}$$ where $$g(z)=f^{\tau}(z) \neq \infty$$ for the minimal value of radius $$R_{ \min}$$ such that $$0 < |z| < \infty$$.
Therefore, let $$g(0)=f^0(0)=0$$ and $$f(g(z))=f^{\tau+1}(z) = \infty$$.
On the other hand, if $$f(z)$$ and $$g(z)$$ are convergent in the complex plane except for infinity, $$f(g(z))$$ must also be convergent in the complex plane except for infinity.
$$\blacksquare$$