My Math Forum Recurrence Relation and Closed Form Relation

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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

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 September 8th, 2014, 09:20 AM #2 Newbie   Joined: Sep 2014 From: LA Posts: 3 Thanks: 1 Cmonnnn guys, no one?
September 8th, 2014, 09:30 AM   #3
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Quote:
 Originally Posted by uniquegel 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.

 September 8th, 2014, 09:34 AM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs 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? Thanks from uniquegel
 September 8th, 2014, 04:18 PM #5 Newbie   Joined: Sep 2014 From: LA Posts: 3 Thanks: 1 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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