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September 7th, 2014, 11:44 PM   #1
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Recurrence Relation and Closed Form Relation

See the attached pic.
I am stuck in Part C).
I figured out the recurrence relation, but couldnt seem to figure out the closed form relation...
I think the R.R should be: Sn = Sn-1+ (3 * 4 ^(n-2))^2 * sqrt(3)/4

Thanks for your helps in advance!
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September 8th, 2014, 09:20 AM   #2
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Cmonnnn guys, no one?
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September 8th, 2014, 09:30 AM   #3
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Originally Posted by uniquegel View Post
Cmonnnn guys, no one?
I saw this thread this morning as I was on my way out, and I found the image to be too small to read on my monitor.
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September 8th, 2014, 09:34 AM   #4
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Okay, I could see it is concerning the Koch snowflake, and this is a response to a problem I posted on another board concerning this:

(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.

Does this help?
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September 8th, 2014, 04:18 PM   #5
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Holy shit man... I LOVE YOU!!! Your like the math superman!!
Unfortunately I just turned my homework in 5 minutes ago...
Apparently a lot of kids in my class didnt figure out Part c)...
Anyways thank you so much, much appreciated!
Thanks from MarkFL
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