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1 change: 1 addition & 0 deletions PMATH370/figures-cache/cobweb-26.md5
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\def \tikzexternallastkey {3D43B2100354074698CE67826ED7EC03}%
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98 changes: 97 additions & 1 deletion PMATH370/notes.tex
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Expand Up @@ -2439,8 +2439,8 @@ \section{Generated iterated function systems}
\end{prf}

\chapter{Complex Functions}
\refstepcounter{section}

\section{Foundations}
\begin{defn*}[complex derivative]
Let $f : \C \to \C$. Then,
\begin{enumerate}
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Then, $H(z) = H(e^{i\theta}) = e^{i\theta} + e^{-i\theta} = 2\cos\theta \in [-2,2]$.

Therefore, $H$ is well-behaved (i.e., invertible) on $R \to \C \setminus [-2,2]$.

\lecture{Mar 13}
Consider now $H(Q_0(z)) = H(z^2) = z^2 + \frac{1}{z^2}$.
Note that $Q_{-2}(H(z)) = (z+\frac{1}{z})^2 - 2 = z^2+\frac{1}{z^2}$.
Hence, $H(Q_0^n(z)) = Q_{-2}^n(H(z))$.

This looks quite similar to $S(Q_c^n(x)) = \sigma^n(S(x))$ in $\R$.
We can say that $H$ plays a similar role as $S$.
In fact, (not course content), $Q_0$ and $Q_{-2}$ are \term*{conjugate}
because $H$ is a homeomorphism between them.

Let $z_n$ be a diverging sequence $\abs{z_n} \to \infty$.
Note that $\abs{H(z_n)} = \abs{z_n + \frac{1}{z_n}} \geq \abs{z_n} - \frac{1}{\abs{z_n}} \to \infty$
Therefore, the image of the sequence $|H(z_n)| \to \infty$ also diverges.

Let $z \in \C \setminus [-2, 2]$.
Since $H$ is surjective, we know there exists a $w \in R$ such that $z=H(w)$,
and see that
\[
\abs{Q_{-2}^n(z)}
= \abs{Q_{-2}^n(H(w))}
= \abs*{H\underbrace{(Q_0^n(w))}_{\to \infty}} \to \infty
\]
by the previous claim.
Hence, $z \not\in K_{-2}$ and we have that $K_{-2} \subseteq [-2,2]$.

Finally, let $z \in [-2, 2]$.
By graphical analysis,
\begin{center}
\cobweb[1.7][domain=-2:2,ymin=-2,ymax=2]{(\x)^2 - 2}{25}
\end{center}
there is no way to escape the box.
That is, $z \in K_{-2}$, i.e., $[-2,2] \subseteq K_{-2}$.

Therefore, $K_{-2} = [-2, 2]$, and we have that $J_{-2} =[-2,2]$.
\end{sol}

\begin{prop}[Escape Criterion]\label{prop:esc}
If $\abs{z} \geq \abs{c} > 2$, then $\abs{Q_c^n(z)} \to \infty$.
In particular, $z \not\in K_c$.
\end{prop}
\begin{prf}
We can write
\[ \abs{Q_c(z)} = \abs{z^2+c} \geq \abs{z}^2 - \abs{c} \geq \abs{z}^2 - \abs{z} = \abs{z}(\abs{z}-1) \]
Suppose $\abs{z} > 2 + \lambda$ for some $\lambda > 0$.
Then, we have that $\abs{z} - 1 > 1+\lambda$.
Therefore, $\abs{Q_c(z)} \geq \abs{z}(1+\lambda)$.

Iterating, we see that $\abs{Q_c^n(z)} \geq \abs{z}(1+\lambda)^n \to \infty$.
\end{prf}

\begin{corollary}
Suppose $\abs{c} > 2$. Then, $\abs{Q_c^n(0)} \to \infty$ and $0 \not\in K_c$.
\end{corollary}
\begin{prf}
Let $z = Q_c(0) = c$ and $\abs{z} = \abs{c} > 2$.
By the \nameref{prop:esc}, $|Q_c^n(0)| \to \infty$.
\end{prf}

\begin{corollary}
Let $M = \max\{\abs{c}, 2\}$.
If $\abs{z} > M$, then $\abs{Q_c^n(z)} \to \infty$.
That is, we have that $K_c \subseteq \{z : \abs{z} \leq M\}$.
\end{corollary}
\begin{prf}
We have $\abs{Q_c^n(z)} \geq (1+\lambda)^n\abs{z} \to \infty$
by the proof of the Escape Criterion
(not the Escape Criterion itself because we don't know if $\abs{z} < 2$).
\end{prf}

\begin{remark}[assignment hint!]
The fact that $K_c$ is inside this bounded disc will help with the proof of its closedness.
\end{remark}

\begin{corollary}
If there exists a $k$ such that $\abs{Q_c^k(z)} > \max\{\abs{c}, 2\}$,
then $\abs{Q_c^n(z)} \to \infty$.
That is, $z \not\in K_c$.
\end{corollary}

Based on these results, we can develop the

\begin{algorithm}[H]
\caption{Filled Julia set algorithm}
\begin{algorithmic}[1]
\State Choose a large $N \in \N$.
\For{points $z$}
\If{$\abs{Q_c^i(z)} > \max\{\abs{c},2\}$ for any $i \leq N$}
\State Colour $z$ white
\ElsIf{$\abs{Q_c^i(z)} \leq \max\{\abs{c},2\}$ for all $i \leq N$}
\State Colour $z$ black
\EndIf
\EndFor
\end{algorithmic}
\end{algorithm}

whose black-shaded region approximates $K_c$.

\pagebreak
\phantomsection\addcontentsline{toc}{chapter}{Back Matter}
\renewcommand{\listtheoremname}{List of Named Results}
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2 changes: 2 additions & 0 deletions latex/agony-pmath370.tex
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Expand Up @@ -27,6 +27,8 @@

\usetikzlibrary{hobby,arrows.meta,decorations.fractals,lindenmayersystems}
\RequirePackage{pstricks,pst-fractal} % for Sierpinski
\RequirePackage{algorithm,float}
\RequirePackage[noEnd]{algpseudocodex}

\pgfdeclarelindenmayersystem{box fractal}{
\rule{F -> F+F-F-F+F}
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