The limit of the perimeter of a Koch snowflake as s(0) goes to zero

In summary, the conversation discussed the perimeter of a Koch snowflake iteration, which tends to infinity as the initial side length, s, tends to zero. This is demonstrated by the limits of ##\lim_{n \to \infty} \left(3s\cdot \left(\dfrac{4}{3}\right)^n\right)=\infty ## and ##\lim_{s \to 0} \left(3s\cdot \left(\dfrac{4}{3}\right)^n\right)=0. ## The question also raised the importance of considering the order of limits and whether the variables are independent or dependent. It was suggested to use the iterative equation for the perimeter to demonstrate the limit
  • #1
nomadreid
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Letting the original triangle of the iteration that results in the Koch snowflake =s: Taking the limit of the Koch snowflake perimeter as s goes to zero intuitively would be infinity, although at s=0 the perimeter would be zero. I'm not sure how that works as an iterated limit.
On one side, if I have any finite value of s = the side of the original triangle of the Koch snowflake iteration, then the perimeter is infinite, so intuitively
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On the other hand, if I looked at the end result first and considered how it got there, then intuitively
BBBB.PNG

(Obviously at n=infinity and s=0, the perimeter is zero, but I am wondering about the limits)

These are the intuitive answers, but intuition is a lousy guide, so I am not sure how this works out as an iterated limit.
Thanks for any pointers.

(Note: this is not a homework question or part of any course, so if the problem is not well posed, I am the clumsy one.)
 
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  • #2
##\lim_{n \to \infty} \left(3s\cdot \left(\dfrac{4}{3}\right)^n\right)=\infty ## and ##\lim_{s \to 0} \left(3s\cdot \left(\dfrac{4}{3}\right)^n\right)=0. ## Now, do whatever is allowed to ##\infty ## and ##0##, e.g. writing ##\lim_{T \to C}## in front of it.What do you really want to know?
 
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  • #3
fresh_42 said:
##\lim_{n \to \infty} \left(3s\cdot \left(\dfrac{4}{3}\right)^n\right)=\infty ## and ##\lim_{s \to 0} \left(3s\cdot \left(\dfrac{4}{3}\right)^n\right)=0. ## Now, do whatever is allowed to ##\infty ## and ##0##, e.g. writing ##\lim_{T \to C}## in front of it.What do you really want to know?
First, I am not sure whether your "T" and "C" are just repeating what I wrote, being n, s, 0, infinity... and what you mean by "whatever is allowed". What I would like to know is what would be the area of the Koch snowflake as s tends to zero--- infinity (my guess) or zero. Which order of the limits is the correct way to look at it?
 
  • #4
The limits are what you have written. They are real numbers, resp. infinite big. Putting another limit in front of it doesn't change that. Limits are no procedures. If you want to establish a procedure, then we need to know what ##n\longmapsto s(n)## or likewise ##s\longmapsto n(s)## is.
 
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  • #5
fresh_42 said:
The limits are what you have written. They are real numbers, resp. infinite big. Putting another limit in front of it doesn't change that. Limits are no procedures. If you want to establish a procedure, then we need to know what ##n\longmapsto s(n)## or likewise ##s\longmapsto n(s)## is.
Thanks. If I understand what you are getting at, the answer will depend on whether I treat s and n as independent variables, which order I wish to approach the limit, and if they are dependent variables, what the relationship between s and n would be. This makes sense. Thanks again.
 
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  • #6
My gut feeling $$\infty$$ is the correct limit. It reminded me of a problem I worked on with an approximate delta function near zero, where the sequence to zero at every point, but the integral remained positive.
 
  • #7
mathman said:
My gut feeling $$\infty$$ is the correct limit. It reminded me of a problem I worked on with an approximate delta function near zero, where the sequence to zero at every point, but the integral remained positive.
Only in the case ##s>0##. No sidelength, no perimeter. It is important to avoid any impression of a limit as a procedure. This is wrong. A limit is a unique, single element of ##\mathbb{R}\cup \{\pm\infty \}.##
 
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  • #8
Maybe, at least to get a feel, you can use the Log Limit? Or try for specific values of s that may be easy to compute/intuit, like ##s=1/n## Edit: I think @mathman may be right. Reminds me of the whole( paraphrase) " Coastline of Britain has infinite length) topic, where when you make your resolution/yardstick increasingly smaller, the perimeter goes to ##\infty ##.
 
  • #9
Since the Koch curve is a fractal it may give a good context to also look at how limits are used for calculating fractal dimension rather than "just" settling with integer dimension limits that goes towards infinity or zero.
 
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  • #10
There is some confusion here: ## s ## is the initial side length when ## n = 0 ##, it does not vary with ## n ##. So we don't need to consider ## \lim_{s \to 0} ##, we can just plug ## s = 0 ## into the equation and every term becomes 0.

Alternatively, use the iterative equation for the perimeter ## P_{n+1} = \frac 4 3 P_n ##. You can see that if ## s = 0, P_0 = 3 s = 0 ## and every iteration has zero perimeter.
 
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  • #11
pbuk said:
So we don't need to consider lims→0, we can just plug s=0 into the equation and every term becomes 0.
The question was about the limit as one varied the initial side length; that is,
s=1 ⇒limitn→∞perimeter of snowflake =∞
s=1/2 ⇒limitn→∞perimeter of snowflake =∞
s=1/4 ⇒limitn→∞perimeter of snowflake =∞
...
s>0 ⇒limitn→∞perimeter of snowflake =∞

At s=0, of course, the perimeter =0, but no one assumes continuity, so the limit and the value attained need not be the same value.
 

Related to The limit of the perimeter of a Koch snowflake as s(0) goes to zero

1. What is a Koch snowflake?

A Koch snowflake is a mathematical curve that is created by starting with an equilateral triangle and repeatedly replacing each line segment with a smaller equilateral triangle. This process is continued infinitely, resulting in a fractal shape with a finite area but an infinite perimeter.

2. How is the perimeter of a Koch snowflake calculated?

The perimeter of a Koch snowflake is calculated by summing the lengths of all the line segments that make up the shape. As the number of iterations increases, the perimeter approaches a finite limit, known as the limit of the perimeter.

3. What does "s(0) goes to zero" mean in relation to the perimeter of a Koch snowflake?

"s(0) goes to zero" refers to the starting size of the line segments that make up the Koch snowflake. As s(0) gets smaller and approaches zero, the number of iterations needed to create the snowflake increases, resulting in a more complex shape with a larger perimeter.

4. Why is the limit of the perimeter of a Koch snowflake important?

The limit of the perimeter of a Koch snowflake is an important concept in mathematics because it demonstrates the idea of a finite area with an infinite perimeter. It also illustrates the concept of self-similarity, where the shape of the snowflake is repeated at different scales.

5. Can the limit of the perimeter of a Koch snowflake be calculated exactly?

No, the limit of the perimeter of a Koch snowflake cannot be calculated exactly. However, it can be approximated to any desired degree of accuracy by increasing the number of iterations used to create the snowflake.

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