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Problem of the Week #81 - October 14th, 2013

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Chris L T521

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Jan 26, 2012
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Thanks again to those who participated in last week's POTW! Here's this week's problem!

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Problem: To construct the Koch snowflake curve, start with an equilateral triangle with sides of length 1. Step 1 in the construction is to divide each side into three equal parts, construct an equilateral triangle on the middle part, and then delete the middle part (see the figure below). Step 2 is to repeat Step 1 for each side of the resulting polygon. This process is repeated at each succeeding step. The Koch snowflake curve is the curve that results from repeating this process indefinitely.


Figure: The $n=0$, $n=1$, $n=2$ and $n=3$ approximating curves for the Koch snowflake.​

(a) Let $s_n$, $\ell_n$ and $p_n$ represent the number of sides, the length of a side, and the total length of the $n$th approximating curve (the curve obtained after Step $n$ of the construction), respectively. Find formulas for $s_n$, $\ell_n$ and $p_n$.

(b) Show that $p_n\to\infty$ as $n\to\infty$.

(c) Sum an infinite series to find the area enclosed by the Koch snowflake curve.

(Parts (b) and (c) show that the Koch snowflake curve is indefinitely long but encloses only a finite area.)

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

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Jan 26, 2012
995
This week's problem was correctly answered by MarkFL. You can find his solution below.

(a) The number of sides:

We see that as defined, each iteration divides all of the sides from the previous iteration into 4 smaller sides. This leads to the linear difference equation:

\(\displaystyle s_{n+1}=4s_{n}\) where \(\displaystyle s_{0}=3\)

The characteristic root is $r=4$ and so the closed-form is:

\(\displaystyle s_{n}=c_14^n\)

Using the initial condition, we may determine the parameter $c_1$:

\(\displaystyle s_{0}=c_14^0=c_1=3\)

And so we find:

\(\displaystyle s_{n}=3\cdot4^n\)

The length of a side:

We see that as defined, each iteration as given in step 1 of the construction is to divide each side into 3 equal parts, hence the sides are 1/3 the length of the previous iteration. This leads to the difference equation:

\(\displaystyle \ell_{n+1}=\frac{1}{3}\ell_{n}\) where \(\displaystyle \ell_{0}=1\)

The characteristic root is \(\displaystyle r=\frac{1}{3}\) and so the closed-form is:

\(\displaystyle \ell_{n}=c_13^{-n}\)

Using the initial condition, we may determine the parameter $c_1$:

\(\displaystyle \ell_{0}=c_13^{-0}=c_1=1\)

And so we find:

\(\displaystyle \ell_{n}=3^{-n}\)

The total perimeter:

To find the perimeter, we need only take the product of the number of sides and the length of each side, hence:

\(\displaystyle p_n=s_n\cdot\ell_n=\frac{4^n}{3^{n-1}}\)

(b) The perimeter at infinity:

\(\displaystyle L=\lim_{n\to\infty}p_n=\lim_{n\to\infty}\frac{4^n}{3^{n-1}}\)

\(\displaystyle \frac{L}{3}=\lim_{n\to\infty}\left(\frac{4}{3} \right)^n\)

Taking the natural log of both sides, we obtain:

\(\displaystyle \ln\left(\frac{L}{3} \right)=\lim_{n\to\infty}n\ln\left(\frac{4}{3} \right)=\ln\left(\frac{4}{3} \right)\lim_{n\to\infty}n=\infty\)

Converting from logarithmic to exponential form, we find:

\(\displaystyle L=3e^{\infty}=\infty\)

Thus, we conclude that the perimeter grows without bound as the fractal is iterated to infinity.

(c) The area at infinity:

Let $A_{n}$ represent the area of the $n$th curve. To get $A_{n+1}$ we must add the area of $s_{n}$ equilateral triangles whose side lengths are $\ell_{n+1}$. That is:

\(\displaystyle A_{n+1}=A_{n}+s_{n}\left(\frac{\sqrt{3}}{4}\ell_{n+1}^2 \right)\)

\(\displaystyle A_{n+1}=A_{n}+3\cdot4^n\left(\frac{\sqrt{3}}{4} \left(3^{-(n+1)} \right)^2 \right)\)

\(\displaystyle A_{n+1}=A_{n}+3^{-\left(2n+\frac{1}{2} \right)}\cdot4^{n-1}\) where \(\displaystyle A_{0}=\frac{\sqrt{3}}{4}\)

Thus, the total area of the snowflake after $n$ iterations is:

\(\displaystyle A_{n}=\frac{\sqrt{3}}{4}+\frac{3\sqrt{3}}{16}\sum_{k=1}^{n}\left(\frac{4}{9} \right)^k\)

Let:

\(\displaystyle S_n=\sum_{k=1}^{n}\left(\frac{4}{9} \right)^k\)

\(\displaystyle \frac{4}{9}S_n=\sum_{k=1}^{n}\left(\frac{4}{9} \right)^{k+1}=S_n-\frac{4}{9}+\left(\frac{4}{9} \right)^{n+1}\)

\(\displaystyle \frac{5}{9}S_n=\frac{4}{9}-\left(\frac{4}{9} \right)^{n+1}\)

\(\displaystyle S_n=\frac{4}{5}-\frac{9}{5}\left(\frac{4}{9} \right)^{n+1}=\frac{4}{5}\left(1-\left(\frac{4}{9} \right)^n \right)\)

Thus, we obtain:

\(\displaystyle A_{n}=\frac{\sqrt{3}}{4}\left(1+\frac{3}{5}\left(1-\left(\frac{4}{9} \right)^n \right) \right)=\frac{\sqrt{3}}{20}\left(8-\left(\frac{4}{9} \right)^n \right)\)

And as a consequence, we find:

\(\displaystyle A_{\infty}=\lim_{n\to\infty}A_n=\frac{\sqrt{3}}{20}\left(8-0 \right)=\frac{2\sqrt{3}}{5}\)

Thus, we have shown that the Koch snowflake is an infinitely long curve enclosing a finite area.
 
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