S: 1,0.3,0.22,0.12,0.06 N: 7,201,478,2595,17950 (no idea how to put this in a tally)
I've got a question here where S is not shrunk by the same fraction throughout, I know how to work out the dimension Fractal when the Shrink is continuous E.G 1/3 each time but not when it's shrunk different each time. Can anyone help?
thanks
In the definition of box-counting dimension, the dimension $d$ is given by $$d = \lim_{\varepsilon \rightarrow 0} \frac{\log N(\varepsilon)}{\log 1/\varepsilon},$$ where $N(\varepsilon)$ is the smallest number of sets of diameter $\varepsilon$ required to cover the set. In numerical applications, $N(\varepsilon)$ is often estimated using any of a number of equivalent formulations. It sounds like your assignment presents the following estimates:
\begin{align} N(1) &= 7 \\ N(0.3) &= 201 \\ N(0.22) &= 478 \\ N(0.12) &= 2595 \\ N(0.06) &= 17950 \end{align}
It might be tempting to simply use the smallest given value of $\varepsilon$ to conclude that your dimension is best approximated by $$\frac{\log(17950)}{\log(1/0.06)} \approx 3.48.$$ One problem with this is that we would typically like to take this sparse data, given over a broad range of $\varepsilon$ values and extrapolate a limit. In this example, if we compute $\log(N(\varepsilon))/\log(1/\varepsilon))$ for the other given values of $\varepsilon$, we see that the computations are decreasing sharply. Thus, the usual approach is to plot $\log(N(\varepsilon))$ versus $\log(1/\varepsilon)$ like so:
Thus, the $x$ coordinates are \begin{align} \log(1) &= 0 \\ \log(1/0.3) &\approx 1.2 \\ \log(1/0.22) &\approx 1.5 \\ \log(1/0.12) &\approx 2.1 \\ \log(1/0.06) &\approx 2.8 \end{align} The $y$ coordinates are $\log(N(1)), \log(N(0.3)),\ldots,\log(N(0.06))$. The line is a least squares fit to this data. The ridiculously good fit suggests that the problem has been manufactured. Since we're interested in how $\log(N(\varepsilon))$ changes in response to a change of $\log(1/\varepsilon)$, the slope of that line represents the fractal dimension. Any number of techniques can be used to compute the least squares fit to be $y\approx 1.945 + 2.79 x$.