- #1
transmini
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On an exam we just took, we were asked to find the dimension of a set using the box counting technique. So choose an epsilon, and cover your object in boxes of side length epsilon, and count the minimum number of boxes required to cover the object. Then use a smaller epsilon and and count the new minimum number of boxes and find the dimension using $$D=\frac{log(N_{i+1}/N_i)}{log(\epsilon_i/\epsilon_{i+1})}$$.
The problem given was essentially the cantor set, except instead of removing the middle third of each layer, you removed the middle fifth. So if the first segment went from ##[0, 1]## then the second segment was ##[0, \frac{2}{5}]\cup[\frac{3}{5}, 1]## and the third was ##[0, \frac{4}{25}]\cup[\frac{6}{25}, \frac{2}{5}]\cup[\frac{3}{5}, \frac{19}{25}]\cup[\frac{21}{25}, 1]## and so on.
Here's where the issue comes in. This should work no matter the epsilon sequence you choose, but we are getting 2 separate answers.
Using ##\epsilon=\frac{1}{5}## on the third segment requires ##\frac{4*\frac{4}{25}}{\frac{1}{5}} \approx 4## boxes. and ##\epsilon=\frac{1}{25}## on the same segment requires ##\frac{4*\frac{4}{25}}{\frac{1}{25}} = 16## boxes. This would then give ##D = \frac{log(\frac{16}{4})}{log(\frac{1/25}{1/5})} = \frac{log(4)}{log(5)}## which is what most people found, including the professor.
However, using ##\epsilon=\frac{2}{5}## on the third segment requires ##\frac{4*\frac{4}{25}}{\frac{2}{5}} \approx 2## boxes. and ##\epsilon=\frac{4}{25}## on the same segment requires ##\frac{4*\frac{4}{25}}{\frac{4}{25}} = 4## boxes. This would then give ##D = \frac{log(\frac{4}{2})}{log(\frac{4/25}{2/5})} = \frac{log(2)}{log(\frac{5}{2})}## which is what I thought the answer should have been.
Since there is a discrepancy here, there is also a formula using fractals given in the book (however the problem specifically said to use the box counting method) which gives the dimension to be $$D = \frac{log(N)}{log(\epsilon)}$$ but here ##N## is the number of self similar copies and ##\epsilon## is the length of the original relative to the copies (so essentially the factor to scale from the copies to the original). Using THIS method on the third segment, gives 4 copies, each of length ##\frac{4}{25}## relative to the original. So ##N=4## and ##\epsilon=\frac{25}{4}##. Plugging into the formula gives ##D=\frac{log(4)}{log(\frac{25}{4})} = \frac{2*log(2)}{2*log(\frac{5}{2})} = \frac{log(2)}{log(\frac{5}{2})}##. Which matches what I believe the answer should have been.
Ultimately, my question here is why is there this discrepancy. How can using 2 different sequences of ##\epsilon## and ##N(\epsilon)## yield 2 completely different dimensions for the same set? Is there a reason that a particular ##\epsilon##-sequence does not work? I read that having an epsilon smaller than your "particle" or "cell" doesn't contribute new information so it essentially would not work, but using the same ##\epsilon##-sequences on the second segment rather than the third yields the same results.
The problem given was essentially the cantor set, except instead of removing the middle third of each layer, you removed the middle fifth. So if the first segment went from ##[0, 1]## then the second segment was ##[0, \frac{2}{5}]\cup[\frac{3}{5}, 1]## and the third was ##[0, \frac{4}{25}]\cup[\frac{6}{25}, \frac{2}{5}]\cup[\frac{3}{5}, \frac{19}{25}]\cup[\frac{21}{25}, 1]## and so on.
Here's where the issue comes in. This should work no matter the epsilon sequence you choose, but we are getting 2 separate answers.
Using ##\epsilon=\frac{1}{5}## on the third segment requires ##\frac{4*\frac{4}{25}}{\frac{1}{5}} \approx 4## boxes. and ##\epsilon=\frac{1}{25}## on the same segment requires ##\frac{4*\frac{4}{25}}{\frac{1}{25}} = 16## boxes. This would then give ##D = \frac{log(\frac{16}{4})}{log(\frac{1/25}{1/5})} = \frac{log(4)}{log(5)}## which is what most people found, including the professor.
However, using ##\epsilon=\frac{2}{5}## on the third segment requires ##\frac{4*\frac{4}{25}}{\frac{2}{5}} \approx 2## boxes. and ##\epsilon=\frac{4}{25}## on the same segment requires ##\frac{4*\frac{4}{25}}{\frac{4}{25}} = 4## boxes. This would then give ##D = \frac{log(\frac{4}{2})}{log(\frac{4/25}{2/5})} = \frac{log(2)}{log(\frac{5}{2})}## which is what I thought the answer should have been.
Since there is a discrepancy here, there is also a formula using fractals given in the book (however the problem specifically said to use the box counting method) which gives the dimension to be $$D = \frac{log(N)}{log(\epsilon)}$$ but here ##N## is the number of self similar copies and ##\epsilon## is the length of the original relative to the copies (so essentially the factor to scale from the copies to the original). Using THIS method on the third segment, gives 4 copies, each of length ##\frac{4}{25}## relative to the original. So ##N=4## and ##\epsilon=\frac{25}{4}##. Plugging into the formula gives ##D=\frac{log(4)}{log(\frac{25}{4})} = \frac{2*log(2)}{2*log(\frac{5}{2})} = \frac{log(2)}{log(\frac{5}{2})}##. Which matches what I believe the answer should have been.
Ultimately, my question here is why is there this discrepancy. How can using 2 different sequences of ##\epsilon## and ##N(\epsilon)## yield 2 completely different dimensions for the same set? Is there a reason that a particular ##\epsilon##-sequence does not work? I read that having an epsilon smaller than your "particle" or "cell" doesn't contribute new information so it essentially would not work, but using the same ##\epsilon##-sequences on the second segment rather than the third yields the same results.