I have a project in my high school trig class to relate math to real life, or just cool other math. Can any of you guys dumb down the math for Sierpinski's triangle and somehow relate it to logarithims?
KTHX!
Sierpinski's triangle
Moderators: gmalivuk, Moderators General, Prelates
 Xanthir
 My HERO!!!
 Posts: 5426
 Joined: Tue Feb 20, 2007 12:49 am UTC
 Location: The Googleplex
 Contact:
Re: Sierpinski's triangle
(defun fibs (n &optional (a 1) (b 1)) (take n (unfold '+ a b)))
Re: Sierpinski's triangle
There's calculating fractal dimension (Sierpinski's triangle is a fractal, as you probably know) and that definitely involves logarithms.
http://www.vanderbilt.edu/AnS/psycholog ... ctals.html
http://www.vanderbilt.edu/AnS/psycholog ... ctals.html

 Posts: 9
 Joined: Sun Feb 06, 2011 1:29 am UTC
Re: Sierpinski's triangle
Nope still makes little sense....
 jestingrabbit
 Factoids are just Datas that haven't grown up yet
 Posts: 5967
 Joined: Tue Nov 28, 2006 9:50 pm UTC
 Location: Sydney
Re: Sierpinski's triangle
Well, I'll try and explain the motivation behind Hausdorff dimension, and from there, you can maybe see what's going on.
So, say you have a square, with side lengths 1. I give you an unlimited number of squares with side length 1/k, then ask you to cover up the square. The smallest number of little squares that you'll need to cover up that square is [imath]k^2[/imath]. What's more interesting is that if you have an arbitrary shape, like a circle, or a triangle or some sort blobular thing, the number of squares of size 1/k that you'll need will be at least [imath]Ak^2[/imath], where A is the area of the shape. In fact, the area is the largest number that you can use in that role.
Now imagine that you have a straight line, or a nice smooth curve, like a circle, or something like that. How many squares of size 1/k would you need to cover it? You'd need at least [imath]\sqrt{2}Lk[/imath] where L is the length, and again the length is the largest number that has that property.
Now, say you had some demon that you had captured and could command. All the demon can do is draw things, but for you to tell it to make a particular drawing, you need to tell it how much ink it will need if its one dimensional, or paint if its two dimensional. Now, maybe you want it to make a "y" shape. There's just one point where it isn't a simple curve, so you can tell it to use such and such ink and make a y shape. Maybe you want a silhouette of someone, and you tell it how much paint it will need, and its all fine.
But now it comes to the problem of you trying to tell the demon to draw the Sierpinski Triangle. Does it have length or area? Well, if you try and calculate the area using your usual techniques, you'll have a geometric series to work out the area that you delete from the big triangle, and I'll leave it to you to work out that it has no area. On the other hand, if you try to work out the length of the horizontal lines in it, you can work out that there's an infinite amount of length.
So you can't tell that demon to draw the Sierpinski Triangle, because that demon can only do length or area. You need a better demon.
So now, notice that area was about shapes where the number of squares you need to cover is at least [imath]Ak^2[/imath], and where A is positive. Length makes most sense when there is a finite L where the number of squares you need is at least [imath]\sqrt{2}Lk.[/imath] So in general, you could ask about shapes where the number of length k squares you need is at least [imath]Ck^d.[/imath] So, given a particular shape, you can ask
"Given some shape S, what is the smallest d such that for any k there is some constant C, such that you need at least [imath]Ck^d[/imath] squares of side length 1/k to cover it?"
That d is Hausdorff dimension. If you look at the picture here
http://en.wikipedia.org/wiki/File:Sierp ... square.svg
you should be able to write a formula to say how many squares you need of size [imath]1/2^n[/imath] to cover a slightly stretched Sierpinski Triangle (imagine that that first square has side length 1). What is that formula? If you let [imath]k = 2^n[/imath], then can you make it be of the form [imath]Ck^d?[/imath] What would you guess the Hausdorff dimension of the Sierpinski Triangle to be?
If you're not getting that, be specific about where it gets hazy.
So, say you have a square, with side lengths 1. I give you an unlimited number of squares with side length 1/k, then ask you to cover up the square. The smallest number of little squares that you'll need to cover up that square is [imath]k^2[/imath]. What's more interesting is that if you have an arbitrary shape, like a circle, or a triangle or some sort blobular thing, the number of squares of size 1/k that you'll need will be at least [imath]Ak^2[/imath], where A is the area of the shape. In fact, the area is the largest number that you can use in that role.
Now imagine that you have a straight line, or a nice smooth curve, like a circle, or something like that. How many squares of size 1/k would you need to cover it? You'd need at least [imath]\sqrt{2}Lk[/imath] where L is the length, and again the length is the largest number that has that property.
Now, say you had some demon that you had captured and could command. All the demon can do is draw things, but for you to tell it to make a particular drawing, you need to tell it how much ink it will need if its one dimensional, or paint if its two dimensional. Now, maybe you want it to make a "y" shape. There's just one point where it isn't a simple curve, so you can tell it to use such and such ink and make a y shape. Maybe you want a silhouette of someone, and you tell it how much paint it will need, and its all fine.
But now it comes to the problem of you trying to tell the demon to draw the Sierpinski Triangle. Does it have length or area? Well, if you try and calculate the area using your usual techniques, you'll have a geometric series to work out the area that you delete from the big triangle, and I'll leave it to you to work out that it has no area. On the other hand, if you try to work out the length of the horizontal lines in it, you can work out that there's an infinite amount of length.
So you can't tell that demon to draw the Sierpinski Triangle, because that demon can only do length or area. You need a better demon.
So now, notice that area was about shapes where the number of squares you need to cover is at least [imath]Ak^2[/imath], and where A is positive. Length makes most sense when there is a finite L where the number of squares you need is at least [imath]\sqrt{2}Lk.[/imath] So in general, you could ask about shapes where the number of length k squares you need is at least [imath]Ck^d.[/imath] So, given a particular shape, you can ask
"Given some shape S, what is the smallest d such that for any k there is some constant C, such that you need at least [imath]Ck^d[/imath] squares of side length 1/k to cover it?"
That d is Hausdorff dimension. If you look at the picture here
http://en.wikipedia.org/wiki/File:Sierp ... square.svg
you should be able to write a formula to say how many squares you need of size [imath]1/2^n[/imath] to cover a slightly stretched Sierpinski Triangle (imagine that that first square has side length 1). What is that formula? If you let [imath]k = 2^n[/imath], then can you make it be of the form [imath]Ck^d?[/imath] What would you guess the Hausdorff dimension of the Sierpinski Triangle to be?
If you're not getting that, be specific about where it gets hazy.
Last edited by jestingrabbit on Wed Mar 09, 2011 2:28 am UTC, edited 1 time in total.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
Re: Sierpinski's triangle
What doesn't make sense? Did you read the wikipedia article? What parts don't you understand? Why do you want to related this to logarithms?
double epsilon = .0000001;
Who is online
Users browsing this forum: No registered users and 8 guests