For the first question, we need to fill in the blank, if only a few, all, or most chaotic systems have fractal state-space structure
And the answer to this is most
While having fractal state-space structure, or having an attractor with fractal dimension, is often a signature of a chaotic system, it is not the case that all chaotic systems have fractal dimension
Questions 2 and 3 really go together
Some people define fractals as self-similar objects
Others define fractals as objects which have non-integer fractal dimension, in particular non-integer Hausdorff dimension
The problem is, that depending on which of these definitions is described to you, the other definition can sometimes be wrong
That is, there are objects that are self-similar, but do not have a non-integer Hausdorff dimension
And there are objects with non-integer Hausdorff dimensions that are not self-similar
For the purposes of this course, we will consider fractals to be objects that have non-integer fractal dimension, in particular non-integer Hausdorff dimension
For the purposes of this quiz, both of these are true
It is important, however, to keep in mind that, when some people say fractals, they mean self-similar, and when other people say fractals, they mean that the object has a non-integer Hausdorff dimension
While, the vast majority of the time, self-similar objects also have non-integer fractal dimension, and vice versa, there are pathological examples that contradict one way or the other
And this is something to keep in mind
Question 4 asks if the fractal dimension of the middle-fifth-removed Cantor set is smaller than that of the middle-third-removed Cantor set
Since these are both Cantor sets, we know that theyll have a dimension between 0 and 1
That is, theyll be somewhere between a point and a line
To figure out which will have a smaller fractal dimension, we need to think about which would be closer to a point than a line
With the middle-third-removed Cantor set, were removing more mass of the line than in the middle-fifth-removed Cantor set
So the middle-third-removed Cantor set will have a smaller fractal dimension, because it will be more similar to a point than to a line
And the middle-fifth-removed Cantor set will have a larger fractal dimension than the middle-third-removed Cantor set, because the middle-fifth-removed Cantor set has less mass removed, and will thus be closer to a line
Since the middle-fifth-removed Cantor set will have a larger fractal dimension than the middle-third-removed Cantor set, this question is false
For this question, we simply need to remember that log of a to the b is equal to b times log of a
Using this identity, we see that log of 2 to the 3 is equal to 3 times log of 2
Question 6 asks us to calculate the capacity dimension of the middle-third-removed Cantor set
This is actually done step-by-step in lecture, but I recommend, if youve never done this, to do this by hand without using the lecture
And come back to the lectures if you get stuck
Its actually a really good mental exercise
As we saw in lecture, the capacity dimension of the middle-third-removed Cantor set is 0.6309
For the final question, what is the capacity dimension of the middle-fifth-removed Cantor set?
While this procedure is very similar to calculating the middle-third-removed Cantor set, lets go through it step-by-step
For simplicity, lets assume this line is 1 long
For the k = 0 step, we can cover this line with one ball of size epsilon equal to 1
In the k = 1 step, we remove one fifth of this line
That means that the remaining pieces are 2/5 long, and we have 2 of them
So in the case of k = 1, we need N(epsilon) = 2, and epsilon = 2/5
In the k = 2 step, we remove one fifth of something two-fifths long
That means the pieces left are 2/5 of 2/5, or 2/5 squared
And we doubled, because we cut every single one into two pieces
Similarly, in the k = 3 step, the remaining pieces are now two-fifths of 2/5 squared, or 2/5 cubed
And we need twice as many of these, so now we have 2 times 2 squared, or 2 cubed
If you continue in this manner, youll see the general pattern that you need N(epsilon) to be 2 to the k, and epsilon to be at 2/5 to the k
If we then plug this into the capacity dimension equation, we see that the capacity dimension, using the same limit trick as was used in lecture, you get log(2) over log(5/2)
Which equals 0.7565, and that is the capacity dimension of the middle-fifth-removed Cantor set
And as you just saw, the capacity dimension of the middle-fifth-removed Cantor set is 0.7565