Sierpiń́ski's fractal and calculating the total blank space

[joej24]

Sierpiń́ski's fractal and calculating the total "blank" space

Homework Statement

Connect the midpoints of the sides of an equilateral triangle to form 4 smaller
equilateral triangles. Leave the middle small triangle blank, but for each of the
other 3 small triangles, draw lines connecting the midpoints of the sides to create
4 tiny triangles. Again leave each middle tiny triangle blank and draw the lines to
divide the others into 4 parts.

Find the infinite series for the total area left blank if this process is continued indefinitely.
(Suggestion: Let the area of the original triangle be 1; then the area of the first blank triangle is 1/4.)
Sum the series to find the total area left blank. Is the answer what you expect?
Hint: What is the “area”of a straight line?

(Comment: You have constructed a fractal called the Sierpi ́ski gasket.
A fractal has the property that a magnified view of a small part of it looks
very much like the original.)

Homework Equations

$$S \,= \, \frac{a}{1-r}$$

The Attempt at a Solution

I understand that the series that represents the total blank space is $$\frac{1}{3}\, \sum \limits_{k=1}^\infty (\frac{3}{4})^k$$ (assuming the area of the whole triangle is 1).
This simplifies to $$\frac{1}{3} \, \frac{\frac{3}{4}}{1-\frac{3}{4}}$$ or $$1$$.
I don't understand how the total blank space adds up to an area of 1 when the total area of the triangle is 1.

In regards to the "hint" in the problem statement, isn't the area of a line zero?

 

[LCKurtz]

You keep chopping out more and more area which adds to 1 as you have shown. What's left is the lines around the edges of the triangles which, as you have noticed, has area 0.

 

[Dick]

You keep chopping out more and more area which adds to 1 as you have shown. What's left is the lines around the edges of the triangles which, as you have noticed, has area 0.

It's a little oversimplified to say just the 'lines' are left. You should notice in the final figure is made of three copies of the figure with dimensions scaled by 1/2. In general the 'volume' a geometric figure should scale by R^n where n is the dimension of the figure (n=1 for lines, n=2 for surface area). This figure scales by the rule (1/2)^n=(1/3). If you work out n it's log(3)/log(2)=1.58... That's a Hausdorff dimension and the figure is a fractal. It's more interesting than just lines. In fact, if you think it is just a countable union of lines, there's a theorem tells you a countable union of lines should have Hausdorff dimension 1. That means there are MANY more points in the gasket that aren't on any of the lines.

 

[joej24]

In general the 'volume' a geometric figure should scale by R^n where n is the dimension of the figure (n=1 for lines, n=2 for surface area). This figure scales by the rule (1/2)^n=(1/3). If you work out n it's log(3)/log(2)=1.58... That's a Hausdorff dimension and the figure is a fractal. It's more interesting than just lines. In fact, if you think it is just a countable union of lines, there's a theorem tells you a countable union of lines should have Hausdorff dimension 1. That means there are MANY more points in the gasket that aren't on any of the lines.

I have some questions. What does the "volume" the figure should scale by mean? In this case, aren't we concerned with the area? And how is the dimension of the figure not 2 if it's on paper (having a length and a width and being flat)? I'm not familiar with the terms Hausdorff and gasket either.

The blank areas approach zero, but never become just lines correct?

 

[Dick]

I have some questions. What does the "volume" the figure should scale by mean? In this case, aren't we concerned with the area? And how is the dimension of the figure not 2 if it's on paper (having a length and a width and being flat)? I'm not familiar with the terms Hausdorff and gasket either.

The blank areas approach zero, but never become just lines correct?

Yes, the short answer is that the blank areas approach zero area, but they never really become just a collection of lines. Things with zero area actually can actually have a pretty interesting structure beyond just being lines. This is one of them. Sierpinski gasket is just a name for the one you you are dealing with. If you've never heard of Hausdorff dimension that's probably fine, your course may not be going into that level of detail on the subject. The point is that fractal structures like this one can have unusual scaling behavior which puts them in a domain between 'lines' and 'planes'.

 

[EnricoHendro]

Sierpiń́ski's fractal and calculating the total "blank" space

Homework Statement

Connect the midpoints of the sides of an equilateral triangle to form 4 smaller
equilateral triangles. Leave the middle small triangle blank, but for each of the
other 3 small triangles, draw lines connecting the midpoints of the sides to create
4 tiny triangles. Again leave each middle tiny triangle blank and draw the lines to
divide the others into 4 parts.

Find the infinite series for the total area left blank if this process is continued indefinitely.
(Suggestion: Let the area of the original triangle be 1; then the area of the first blank triangle is 1/4.)
Sum the series to find the total area left blank. Is the answer what you expect?
Hint: What is the “area”of a straight line?

(Comment: You have constructed a fractal called the Sierpi ́ski gasket.
A fractal has the property that a magnified view of a small part of it looks
very much like the original.)

Homework Equations

$$S \,= \, \frac{a}{1-r}$$

The Attempt at a Solution

I understand that the series that represents the total blank space is $$\frac{1}{3}\, \sum \limits_{k=1}^\infty (\frac{3}{4})^k$$ (assuming the area of the whole triangle is 1).
This simplifies to $$\frac{1}{3} \, \frac{\frac{3}{4}}{1-\frac{3}{4}}$$ or $$1$$.
I don't understand how the total blank space adds up to an area of 1 when the total area of the triangle is 1.

In regards to the "hint" in the problem statement, isn't the area of a line zero?

Hey there, I have a question, why is the series that represents the total blank space is 1/3 times 3/4 ? I mean I kind of understand the 3/4 part, which is 3 triangles each "produces" 1/4 (blank triangle area). But why do we have 1/3 there?

 

[Dick]

Hey there, I have a question, why is the series that represents the total blank space is 1/3 times 3/4 ? I mean I kind of understand the 3/4 part, which is 3 triangles each "produces" 1/4 (blank triangle area). But why do we have 1/3 there?

Well, I think it would be more natural to write the series as $\frac{1}{4}\, \sum \limits_{k=0}^\infty (\frac{3}{4})^k$. Does that look correct to you? But that's really the same series that the OP wrote. Just write out some terms to verify.

 

[EnricoHendro]

Well, I think it would be more natural to write the series as $\frac{1}{4}\, \sum \limits_{k=0}^\infty (\frac{3}{4})^k$. Does that look correct to you? But that's really the same series that the OP wrote. Just write out some terms to verify.

Well, if the 1/4 means there is 1 blank triangle for every 4 triangle, then yes, 1/4 makes more sense to me. But I mean, if we put 1/4 instead of 1/3, wouldn’t the answer be 3/4 not 1?? I still can’t grasp the logic behind the 1/3 tho

 

[Dick]

Well, if the 1/4 means there is 1 blank triangle for every 4 triangle, then yes, 1/4 makes more sense to me. But I mean, if we put 1/4 instead of 1/3, wouldn’t the answer be 3/4 not 1?? I still can’t grasp the logic behind the 1/3 tho

I didn't worry about why they wrote it the way they did. Pay attention to the start of the sequence. They started with $k=1$, I started with $k=0$. That's the only difference. But they are still the same series.

 

[EnricoHendro]

I didn't worry about why they wrote it the way they did. Pay attention to the start of the sequence. They started with $k=1$, I started with $k=0$. That's the only difference. But they are still the same series.

ah I get it now. Yes indeed they are the same. I figured I used the wrong r, that's why I got it wrong. Thanks man