As it turns out, the Julia set of $f(z) = z^2 + c$ is either "in on piece" or "totally dusty". This leads to the definition of a Mandelbrot set $M$, which is the set of all parameters $c \in \mathbb{C}$ for which such Julia set is connected:

$$\begin{aligned} M = \{ c \in \mathbb{C}: J(z^2 + c) \text{ is connected} \} \end{aligned}$$

Here are two amazing theorems:

$J(z^2 + c)$ is connected if and only if 0 does not belong to $A(\infty)$ !

A complex number $c$ belongs to $M$ if and only if $\lvert f^n(0) \rvert \le 2$ for all $n \ge 1$ !

You can find more info at Analysis of a Complex Kind.

# posted by rot13(Unafba Pune) @ 9:53 PM 0 comments

Given a function $f$, the Julia set of $f$, $J(f)$, is the boundary of $A(\infty)$, where $A(\infty)$ is the "basin of attraction to infinity", ie.

$$\begin{aligned} A(\infty) = \{ z \in \mathbb{C}: f^n(z) \rightarrow \infty \text{ as } n \rightarrow \infty \} \end{aligned}$$

The filled-in Julia set of $f$, $K(f)$, are those $z \in \mathbb{C}$ for which $f^n(z)$ stays bounded:

$$\begin{aligned} K(f) = \{ z \in \mathbb{C}: \{ f^n(z) \} \text{ is bounded}\} \end{aligned}$$

Given $f(z) = z^2 + c$, where $c \in \mathbb{C}$, how can we find $K(f)$ ? Turns out

$$\begin{aligned} R = \frac{1 + \sqrt{1 + 4\lvert c \rvert}}{2} \end{aligned}$$

may help !

Let $z_0 \in \mathbb{C}$. If for some $n > 0$ we have $\lvert f^n(z_0) \rvert$ > R, then $f^n(z_0) \rightarrow \infty$ as $n \rightarrow \infty$ !

That is, $z_0 \in A(\infty)$, so $z_0 \notin K(f)$.

You can find more info at Analysis of a Complex Kind.

# posted by rot13(Unafba Pune) @ 6:56 PM 0 comments

Let

$$\begin{aligned} \varphi(w) = w + \frac{1}{w} \end{aligned}$$

Why is it true that

$$\begin{aligned} \varphi(w) : \{w : \lvert w \rvert > 1\} \rightarrow \mathbb{C} \,\backslash\, [-2,2] \end{aligned}$$

?

If $\lvert w \rvert \ge 2$, $\varphi(w)$ obviously excludes the interval $[-2,2]$.

If $w = 1$, $\varphi(w) = 2$.

Or more generally, if $\lvert w \rvert = 1$ or $w = e^{i\theta}$,

$$\begin{aligned} \varphi(w) &= e^{i\theta} + e^{-i\theta} \\ &= \cos{\theta} + i\sin{\theta} + \cos{\theta} - i\sin{\theta} \\ &= 2\cos{\theta} \\ \end{aligned}$$

$\therefore \,\,0 \le \lvert \varphi(w) \rvert \le 2$ when $\lvert w \rvert = 1$.

Observe that $\displaystyle \left| \varphi(w) \right|$ is monotonic increasing, which means the interval $[-2,2]$ is clearly excluded from $\displaystyle \varphi(w)$ as $\left| w \right| > 1$.

More info at Analysis of a Complex Kind.

# posted by rot13(Unafba Pune) @ 9:34 PM 0 comments

Given

$$\begin{aligned} \varphi^{-1} (f(\varphi(w))) = w^2 = g(w) \\ \end{aligned}$$

What is $f(z)$ in terms of $g$ ?

Let

$$\begin{aligned} z &= \varphi(w)) \\ \varphi^{-1}(z) &= w \\ \varphi^{-1} (f(\varphi(w))) &= g(w) \\ f(\varphi(w)) &= \varphi(g(w)) \\ \therefore \,\, f(z) &= \varphi(g(\varphi^{-1}(z))) \\ \end{aligned}$$

Why is $\displaystyle \varphi^{-1} \circ f \circ \varphi$ interesting ?

It's rather easy to show that the function $\displaystyle p(z) = \varphi^{-1} \circ f \circ \varphi$ behaves the same as $\displaystyle f(z)$ under iteration. In other words,

$$\begin{aligned} p^n = \varphi^{-1} \circ f^n \circ \varphi \end{aligned}$$

So instead of studying $p$, which can be a bit more complicated, the problem can be reduced to the study of $f$; or vice-versa.

In general, given any quadratic polynomials in $\mathbb{C}$

$$\begin{aligned} p(z) = az^2 + bz + d \\ \end{aligned}$$

Let

$$\begin{aligned} f(z) &= z^2 + c \\ c &= ad + \frac{b}{2} - (\frac{b}{2})^2 \\ \varphi(z) &= az + \frac{b}{2} \\ \end{aligned}$$

It can be readily verified that

$$\begin{aligned} p &= \varphi^{-1} \circ f \circ \varphi \\ f &= \varphi \circ p \circ \varphi^{-1} \\ \end{aligned}$$

Furthermore,

$$\begin{aligned} p^n &= \varphi^{-1} \circ f^n \circ \varphi \\ f^n &= \varphi \circ p^n \circ \varphi^{-1} \\ \end{aligned}$$

More info at Analysis of a Complex Kind.

# posted by rot13(Unafba Pune) @ 1:18 AM 0 comments

What is $\displaystyle \lim_{z \rightarrow i} \frac{z^2 + 1}{z - i}$ ?

I think the answer is neat and a little surprising.

You can find more info at Analysis of a Complex Kind.

# posted by rot13(Unafba Pune) @ 8:59 AM 0 comments

$$\begin{aligned} \frac{1}{i} &= i^{-1} = e^{i\frac{-\pi}{2}} = -i \\ \end{aligned}$$

# posted by rot13(Unafba Pune) @ 10:17 PM 0 comments

Suppose

$$\begin{aligned} a_n = f(n) \,\,\,\,\, \text{ where \(f\) is a decreasing and positive function.} \end{aligned}$$

If

$$\begin{aligned} \int_1^\infty f(x)\,dx \,\,\,\, \text{ is finite,} \end{aligned}$$

then

$$\begin{aligned} \int_1^\infty f(x)\,dx \,\,\, \le \,\,\, \sum_{n=1}^\infty a_n \,\,\, \le \,\,\, a_1 + \int_1^\infty f(x)\,dx \end{aligned}$$

!

You can find more information at Calculus Two: Sequences and Series.

# posted by rot13(Unafba Pune) @ 9:35 PM 0 comments

Let

$$\begin{aligned} f(x) &= \sum_{n=0}^\infty a_n x^n \\ g(x) &= \sum_{n=0}^\infty b_n x^n \\ R &= \text{the minimum of their radii of convergence} \\ \end{aligned}$$

Then

$$\begin{aligned} f(x)\,g(x) &= \sum_{n=0}^\infty (\sum_{i=0}^{n}a_i\, b_{n-i}) \, x^n \\ \end{aligned}$$

for $\displaystyle x \in (-R, R)$ !

You can find more information at Calculus Two: Sequences and Series.

# posted by rot13(Unafba Pune) @ 1:53 PM 0 comments

The first three determinants of a 1,3,1 matrix:

$$S_1 = \begin{vmatrix} 3 \end{vmatrix}$$

$$S_2 = \begin{vmatrix} 3 & 1 \\ 1 & 3 \end{vmatrix}$$

$$S_3 = \begin{vmatrix} 3 & 1 & 0 \\ 1 & 3 & 1 \\ 0 & 1 & 3 \end{vmatrix}$$

Show that $S_n$ is the Fibonacci number $F_{2n+2}$ by proving $F_{2n+2} = 3F_{2n} - F_{2n-2}$. Keep using Fibonacci's rule $F_k = F_{k-1} + F_{k-2}$ starting with $k = 2n + 2$.

Attempt:

First, we can see via cofactors that

$$\begin{aligned} S_1 &= 3 = F_4 \\ S_2 &= 9 - 1 = 8 = F_6 \\ S_3 &= 3S_2 - \begin{vmatrix} 1 & 1 \\ 0 & 3 \end{vmatrix} = 3S_2 - S_1 \end{aligned}$$

In general, we can see that

$$\begin{aligned} S_n = 3S_{n-1} - S_{n-2} \end{aligned}$$

Next, setting $k = 2n + 2$,

$$\begin{aligned} F_{2n+2} &= F_{2n+1} + F_{2n} = (F_{2n} + F_{2n-1}) + F_{2n} \\ &= (3F_{2n} - F_{2n}) + F_{2n-1} \\ &= (3F_{2n} - F_{2n-1} - F_{2n-2}) + F_{2n-1} \\ \therefore F_{2n+2} &= 3F_{2n} - F_{2n-2} \\ \end{aligned}$$

Now, if $S_n = F_{2n+2}$, then $S_{n-1} = F_{2(n-1)+2} = F_{2n}$ and similarly $S_{n-2} = F_{2n-2}$. But as demonstrated above this turns out to be the case for $S_1 = F_4$ and $S_2 = F_6$. Therefore, $S_n$ is indeed the Fibonnaci number $F_{2n+2}$. Interestingly, $S_n$ is a sequence that produces every second Fibonnaci number !

# posted by rot13(Unafba Pune) @ 9:10 PM 0 comments