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