Dynamics

From Iterated Functions

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[edit]

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[edit]

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[edit]

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[edit]

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[edit]

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[edit]

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.

Proof by contradiction.

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.

This is a contradiction which completes the proof.

\(\blacksquare\)

  1. Dynamical Recurrance Equation