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mark

(5 pts)

This assignment is a follow up to the Generate your own carpet IFS assignment. Your mission is:

Use the box-counting techniques to compute the box-counting dimension of your personal carpet.

To be clear, you are not asked to compute the similarity dimension, though the two should probably agree. Itâ€™s still not particularly hard, though; the main point is the write up. Be sure to:

• Identify an appropriate sequence (\varepsilon_n) of mesh sizes decreasing to zero,
• Find the corresponding expression for N_{\varepsilon_n}(E), and
• Simplify \displaystyle \frac{\log N_{\varepsilon_n(E)}}{\varepsilon_n} so you can compute the limit.
pikenber

For my (\varepsilon_n) sequence, my values were 3, 1, \frac{1}{3}, \frac{1}{9},..., which turns into the sequence (\frac{1}{3})^{n-2} = 9(\frac{1}{3})^n, where n\in N.
My corresponding N_{\varepsilon} (E) was 1, 7, 49, 343,..., which can be written as (7)^{n-1} = \frac{1}{7}(7)^n, with n\in N.
Finally, I took \lim_{n\to \infty} \frac{log(\frac{1}{7}(7)^n)}{log(9(\frac{1}{3})^n)} = \lim_{n\to \infty} \frac{log(\frac{1}{7})+nlog(7)}{log(9)+nlog(\frac{1}{3})}. This simplifies as n\to \infty as the constants no longer matter. Thus \lim_{n\to \infty} \frac{nlog(7)}{nlog(\frac{1}{3})} = \lim_{n\to \infty} \frac{log(7)}{log(\frac{1}{3})} \approx -1.77 = -d. I did check the similarity dimension, and it is the same.

ofeldman

For my (\varepsilon_n) sequence, my values were 3, 1, \frac{1}{3}, \frac{1}{9},..., which turns into the sequence (\frac{1}{3})^{n-2} = 9(\frac{1}{3})^n , where n\in N.
My corresponding N_{\varepsilon} (E) was 1, 4, 16, 64,..., which can be written as (4)^{n-1} = \frac{1}{4}(4)^n , with n\in N.
Finally, I took \lim_{n\to \infty} \frac{log(\frac{1}{4}(4)^n)}{log(9(\frac{1}{3})^n)} = \lim_{n\to \infty} \frac{log(\frac{1}{4})+nlog(4)}{log(9)+nlog(\frac{1}{3})}. This simplifies as n\to \infty as the constants no longer matter. Thus \lim_{n\to \infty} \frac{nlog(4)}{nlog(\frac{1}{3})} = \lim_{n\to \infty} \frac{log(4)}{log(\frac{1}{3})} \approx -1.26 = -d. I also checked the similarity dimension, \frac{log(4)}{log(1/3)} \approx -1.26 = -d, and it is the same.

scowart

Given my carpet IFS, I observed my (\varepsilon_n) to follow the pattern \varepsilon_n=3,1/3,1/9,1/27,... or \varepsilon_n=(1/3)^{n-2} such that n\in\mathbb{N}. It can shown that \lim\varepsilon_n=0. Then I determined N_{\varepsilon_n}=1,7,49,343,.. or N_{\varepsilon_n}=7^{n-1}. Combining these results yields

\frac{\log N_{\varepsilon_n}(E)}{\log\varepsilon_n}=\frac{\log(7^{n-1})}{\log((\frac{1}{3})^{n-2})}.

Which can be simplified to

\frac{\log(7^{n-1})}{\log((\frac{1}{3})^{n-2})}=\frac{(n-1)\log(7)}{(n-2)\log(\frac{1}{3})}
=\frac{n\log(7)-\log(7)}{n\log(\frac{1}{3})-2\log(\frac{1}{3})}.

Computing the limit shows

\lim_{n\rightarrow\infty}\frac{n\log(7)-\log(7)}{n\log(\frac{1}{3})-2\log(\frac{1}{3})}=\lim_{n\rightarrow\infty}\frac{n\log(7)}{n\log(\frac{1}{3})}
=\lim_{n\rightarrow\infty}\frac{\log(7)}{\log(\frac{1}{3})}=\frac{\log(7)}{\log(\frac{1}{3})}
\approx-1.77=-d.

Given my IFS, my (\varepsilon_n) sequence followed the pattern \varepsilon_n=3,1, \frac{1}{3},\frac{1}{9},\frac{1}{27}.... This can be written as (\frac{1}{3})^{n-2}=9(\frac{1}{3})^n, where n \in N.
My N_\varepsilon(E) expression was 1,7,49,343,...
This sequence can be written as (7)^{n-1}=\frac{1}{7}(7)^n, again where n \in N.
Taking the limit, \lim_{n \to \infty} \frac{\log(\frac{1}{7}(7)^n)}{log(9(\frac{1}{3})^n)}= \lim_{n \to \infty} \frac{\log(\frac{1}{7}) + n\log(7)}{\log(9) +n\log(\frac{1}{3})}.
As n \to \infty the constants loose their significance. This means \lim_{n \to \infty} \frac{n\log(7)}{n\log(\frac{1}{3})}= \lim_{n \to \infty} \frac{\log(7)}{\log(\frac{1}{3})}\approx -1.77= -d.

lee7

For my carpet IFS, my sequence of \varepsilon_n was \varepsilon_n =3,1, \frac{1}{3}, \frac{1}{9}, \frac{1}{27}... or \varepsilon_n=\frac{1}{3}^{n-2}=9(\frac{1}{3})^n for n\in\mathbb{N}.
My N_\varepsilon(E) sequence was 1, 5, 25, 125... or N_\varepsilon(E) = (5)^{n-1}=\frac{1}{5}(5)^n

The box counting dimension is then given by:
\lim_{n\to\infty} \frac{\log N_{\varepsilon_n}(E)}{\log\varepsilon_n} = \lim_{n\to\infty} \frac{\log \frac{1}{5}(5)^n }{\log9(\frac{1}{3})^n}
The constants, \frac{1}{5} and 9, lose their significance as n \to \infty. Thus
\lim_{n\to\infty}\frac{\log(5)}{\log(\frac{1}{3})}=\frac{\log(5)}{\log(\frac{1}{3})} \approx -1.46=-d.

mark