Filling out the fractal

Yesterday, class was spent discussing how a fractal was filled. This was a very interesting lecture that helped me understand more how fractals will take shape and what they are composed of. Using an algorithm we were introduced to in the class prior, namely:

T' = T_1(A) U T_2(A) ; where A is a set such that
A = {x}

The iterations look something like this

{x} -> {T_1(x), T_2(x)} -> {(T_1)^2(x), T_1T_2(x), (T_2)^2(x)} ->...

and the transformations look like this

T_1(x,y) = |.5 .5, 0 .5| (x, y)


T_2(x, y) = |.5 .5, 0 .5| (x, y) + (0, .5)


At first we tried to perform these transformations on the unit square. This didn't work out because the first iteration of the transformation left no gap in between the subsequent skewed graphs , which as an effect caused the iterations to become static. We then decided to start with a square starting at the origin , going out 2 units in each direction. This caused the first transformation to have a small gap between the 2 parallelograms (the result of the first interation). This allows us to iterate infinitely.

At this point in the lecture we began discussing the idea of a point's address. That is, any point which is an element of the fractal can be assigned a distinct address. An address is a binary string of numbers that you can disect to find out exactly where this point is inside our fractal. For the specific fractal we were looking at, the first index of the address is assigned a 0 if it is on the bottom half of the original square or a one if it is on the top half. The second index of the address is assigned a 0 if it is on the bottom half of the parallelogram created for the first iteration of the transformation, and a 1 if it is on the top half. This is repeated a finite amount of times until you get the address of the exact point you are looking at.

The concept of address will be a little easier to understand if I can post the graph of the fractal we were examaning. As soon as I figure out how to graph it in mathematica, I 'll post the image.

Contracting Mapping

The better part of today's class was spent proving a theorem, which after a little bit of search I found is called the "Banach Fixed Point Theorem".

Banach Fixed Point Theorem:

Let (Y,m) me a complete metric space. Let T: Y -> Y be a continuous map such that for all x,y that are elements of Y, m(Tx,Ty) <= km(x,y) for some |k|<1.> x'.

From Mathworld:

Let be a contraction mapping from a closed subset of a Banach space into . Then there exists a unique such that .

My take on this:

It is easier for me to understand the theorem given to us in class rather than the theorem I found on Mathworld. (Maybe because I have no idea what "Banach space" is.) The way I see this is thatt T is a contracting transformation (which is the reason 0<1) style="font-weight: bold;">Summation of the Proof:

We proved this in 2 parts:

1. There is only 1 fixed point. This was proved by assuming there are 2 and showing they are equal. To do this we had to assume the sequence {(T^n)(y)} is cauchy.
   
2. Prove that the previous sequence is cauchy.

Convergent Series

Back on the 10th, we were given the assignment to find the value of 3 different sums (series).

The first of which was the sum of (1/(2^i)) as i goes from zero to infinity. The number that this series will eventually converge to is 2.

Next we had to find the sum of (1/(2^i)) as i goes from 1 to infinity. This converges to 1.

The last series we were assigned to evalute was the sum of (1/(2^i)) as i goes from 2 to infinity. As far as I can see this should converge to 1/2.

Proof:

1)

The n-th partial sum for this series is defined as
S _n = 1/2 + 1/2 ² + 1/2 ³ + ... + 1/2 n

If we divide the above expression by 2 and then subtract it from the original one we get:

S _n - 1/2 S _n = 1/2 - 1/2 ⁿ⁺¹

Hence, solving this for S _n we obtain

S _n = 2 (1/2 - 1/2 ⁿ⁺¹)

This is now a sequence, and we can take the limit as n goes to infinity. By our result on the power sequence, the term 1/2 ⁿ⁺¹ goes to zero,
so that lim S _n = 1

Taken From:
Interactive Real Analysis, ver. 1.9.4
(c) 1994-2005, Bert G. Wachsmuth

2) For the same series but with n starting at 0 or 2 we can just add or subtract the singleton element from our series limit. i.e. if we start with n=0, then one more element will be added to the series. The element 1/(2^0) = 1, so the limit as n goes to infinity is 1+1=2. On the other hand if we start with one less element such that n=2, then we can subtract the element 1/(2^1)=1/2, thus our limit would be 1-1/2=1/2.