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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
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And the answer to this is most
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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
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Questions 2 and 3 really go together
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Some people define fractals as self-similar objects
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Others define fractals as objects which have non-integer fractal dimension, in particular non-integer Hausdorff dimension
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The problem is, that depending on which of these definitions is described to you, the other definition can sometimes be wrong
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That is, there are objects that are self-similar, but do not have a non-integer Hausdorff dimension
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And there are objects with non-integer Hausdorff dimensions that are not self-similar
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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
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For the purposes of this quiz, both of these are true
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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
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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
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And this is something to keep in mind
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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
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Since these are both Cantor sets, we know that theyll have a dimension between 0 and 1
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That is, theyll be somewhere between a point and a line
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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
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With the middle-third-removed Cantor set, were removing more mass of the line than in the middle-fifth-removed Cantor set
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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
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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
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Since the middle-fifth-removed Cantor set will have a larger fractal dimension than the middle-third-removed Cantor set, this question is false
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For this question, we simply need to remember that log of a to the b is equal to b times log of a
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Using this identity, we see that log of 2 to the 3 is equal to 3 times log of 2
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Question 6 asks us to calculate the capacity dimension of the middle-third-removed Cantor set
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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
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And come back to the lectures if you get stuck
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Its actually a really good mental exercise
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As we saw in lecture, the capacity dimension of the middle-third-removed Cantor set is 0.6309
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For the final question, what is the capacity dimension of the middle-fifth-removed Cantor set?
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While this procedure is very similar to calculating the middle-third-removed Cantor set, lets go through it step-by-step
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For simplicity, lets assume this line is 1 long
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For the k = 0 step, we can cover this line with one ball of size epsilon equal to 1
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In the k = 1 step, we remove one fifth of this line
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That means that the remaining pieces are 2/5 long, and we have 2 of them
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So in the case of k = 1, we need N(epsilon) = 2, and epsilon = 2/5
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In the k = 2 step, we remove one fifth of something two-fifths long
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That means the pieces left are 2/5 of 2/5, or 2/5 squared
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And we doubled, because we cut every single one into two pieces
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Similarly, in the k = 3 step, the remaining pieces are now two-fifths of 2/5 squared, or 2/5 cubed
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And we need twice as many of these, so now we have 2 times 2 squared, or 2 cubed
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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
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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)
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Which equals 0.7565, and that is the capacity dimension of the middle-fifth-removed Cantor set
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And as you just saw, the capacity dimension of the middle-fifth-removed Cantor set is 0.7565