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.

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


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 \(t \in \mathbb{N^+}\).


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 \(t \in \mathbb{C}\).


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 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.


  1. Dynamical Recurrance Equation