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