Total “length” of the Sierpinski triangle
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From the Wikipedia page for the Sierpinski triangle:
The Sierpinski triangle may be constructed from an equilateral triangle by repeated removal of triangular subsets:
- Start with an equilateral triangle.
- Subdivide it into four smaller congruent equilateral triangles and remove the central triangle.
- Repeat step 2 with each of the remaining smaller triangles forever.
Each removed triangle (a trema) is topologically an open set.
So the Sierpinski triangle contains no interior point. If it were not the case, each interior point is must also contained in an open triangle of step 2 and 3 and thus be removed. A contradiction.
So we can consider the Sierpinski triangle consisting the "limiting" result of union line segments (boundaries) which become shorter and shorter "forever".
Question: If we sum up the lengths of the infinitely many line segments (boundaries) consisting the Sierpinski triangle, will it converge? If yes, what is the length compared with the length of edge of the original trianghe
sequences-and-series limits convergence fractals
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From the Wikipedia page for the Sierpinski triangle:
The Sierpinski triangle may be constructed from an equilateral triangle by repeated removal of triangular subsets:
- Start with an equilateral triangle.
- Subdivide it into four smaller congruent equilateral triangles and remove the central triangle.
- Repeat step 2 with each of the remaining smaller triangles forever.
Each removed triangle (a trema) is topologically an open set.
So the Sierpinski triangle contains no interior point. If it were not the case, each interior point is must also contained in an open triangle of step 2 and 3 and thus be removed. A contradiction.
So we can consider the Sierpinski triangle consisting the "limiting" result of union line segments (boundaries) which become shorter and shorter "forever".
Question: If we sum up the lengths of the infinitely many line segments (boundaries) consisting the Sierpinski triangle, will it converge? If yes, what is the length compared with the length of edge of the original trianghe
sequences-and-series limits convergence fractals
2
Obviously it diverges (as the sum of a geometric series with ratio $3/2$).
– metamorphy
Nov 23 at 4:00
add a comment |
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1
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up vote
1
down vote
favorite
From the Wikipedia page for the Sierpinski triangle:
The Sierpinski triangle may be constructed from an equilateral triangle by repeated removal of triangular subsets:
- Start with an equilateral triangle.
- Subdivide it into four smaller congruent equilateral triangles and remove the central triangle.
- Repeat step 2 with each of the remaining smaller triangles forever.
Each removed triangle (a trema) is topologically an open set.
So the Sierpinski triangle contains no interior point. If it were not the case, each interior point is must also contained in an open triangle of step 2 and 3 and thus be removed. A contradiction.
So we can consider the Sierpinski triangle consisting the "limiting" result of union line segments (boundaries) which become shorter and shorter "forever".
Question: If we sum up the lengths of the infinitely many line segments (boundaries) consisting the Sierpinski triangle, will it converge? If yes, what is the length compared with the length of edge of the original trianghe
sequences-and-series limits convergence fractals
From the Wikipedia page for the Sierpinski triangle:
The Sierpinski triangle may be constructed from an equilateral triangle by repeated removal of triangular subsets:
- Start with an equilateral triangle.
- Subdivide it into four smaller congruent equilateral triangles and remove the central triangle.
- Repeat step 2 with each of the remaining smaller triangles forever.
Each removed triangle (a trema) is topologically an open set.
So the Sierpinski triangle contains no interior point. If it were not the case, each interior point is must also contained in an open triangle of step 2 and 3 and thus be removed. A contradiction.
So we can consider the Sierpinski triangle consisting the "limiting" result of union line segments (boundaries) which become shorter and shorter "forever".
Question: If we sum up the lengths of the infinitely many line segments (boundaries) consisting the Sierpinski triangle, will it converge? If yes, what is the length compared with the length of edge of the original trianghe
sequences-and-series limits convergence fractals
sequences-and-series limits convergence fractals
edited Nov 23 at 15:34
Jean-Claude Arbaut
14.6k63363
14.6k63363
asked Nov 23 at 3:54
JZH
727
727
2
Obviously it diverges (as the sum of a geometric series with ratio $3/2$).
– metamorphy
Nov 23 at 4:00
add a comment |
2
Obviously it diverges (as the sum of a geometric series with ratio $3/2$).
– metamorphy
Nov 23 at 4:00
2
2
Obviously it diverges (as the sum of a geometric series with ratio $3/2$).
– metamorphy
Nov 23 at 4:00
Obviously it diverges (as the sum of a geometric series with ratio $3/2$).
– metamorphy
Nov 23 at 4:00
add a comment |
1 Answer
1
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2
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An alternative construction of the Sierpinski gasket (one that is useful if you want to study differential equations in that setting, for example–see Differential Equations on Fractals by Robert S. Strichartz) is to build it as the limit of a sequence of graphs (a graph is a collection of nodes and edges; you can think of it as a network of cities (nodes) joined by roads (edges)). This is a more direct way of discussing the boundary segments from the original question.
To construct the Sierpinski gasket, start with a graph with three nodes and three edges, arranged to form an equilateral triangle (in graph theory, the arrangement of the nodes isn't really important, but I'll impose that structure here, as it makes it easier to think about what is going on). Place a new node in the middle of each of the original nodes, then join them by edges to form the appropriate triangles. Do this again. And again. And again. In the limit, we end up with the Sierpinski gasket. The first three stages are pictured below:
In the $k$-th stages of this process, we add $3^k$ edges, each of which has length $(frac{1}{2})^k$. Thus the total length of the edges added to the original graph can be computed as
$$ sum_{j=1}^{infty} 3^k left(frac{1}{2}right)^k
= sum_{j=1}^{infty} left(frac{3}{2}right)^k, $$
which is a series diverging to infinity. Since the edges added through this process are dense in the limiting gasket, we can reasonably conclude that the "length" of the Sierpinski gasket is infinite.
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
An alternative construction of the Sierpinski gasket (one that is useful if you want to study differential equations in that setting, for example–see Differential Equations on Fractals by Robert S. Strichartz) is to build it as the limit of a sequence of graphs (a graph is a collection of nodes and edges; you can think of it as a network of cities (nodes) joined by roads (edges)). This is a more direct way of discussing the boundary segments from the original question.
To construct the Sierpinski gasket, start with a graph with three nodes and three edges, arranged to form an equilateral triangle (in graph theory, the arrangement of the nodes isn't really important, but I'll impose that structure here, as it makes it easier to think about what is going on). Place a new node in the middle of each of the original nodes, then join them by edges to form the appropriate triangles. Do this again. And again. And again. In the limit, we end up with the Sierpinski gasket. The first three stages are pictured below:
In the $k$-th stages of this process, we add $3^k$ edges, each of which has length $(frac{1}{2})^k$. Thus the total length of the edges added to the original graph can be computed as
$$ sum_{j=1}^{infty} 3^k left(frac{1}{2}right)^k
= sum_{j=1}^{infty} left(frac{3}{2}right)^k, $$
which is a series diverging to infinity. Since the edges added through this process are dense in the limiting gasket, we can reasonably conclude that the "length" of the Sierpinski gasket is infinite.
add a comment |
up vote
2
down vote
accepted
An alternative construction of the Sierpinski gasket (one that is useful if you want to study differential equations in that setting, for example–see Differential Equations on Fractals by Robert S. Strichartz) is to build it as the limit of a sequence of graphs (a graph is a collection of nodes and edges; you can think of it as a network of cities (nodes) joined by roads (edges)). This is a more direct way of discussing the boundary segments from the original question.
To construct the Sierpinski gasket, start with a graph with three nodes and three edges, arranged to form an equilateral triangle (in graph theory, the arrangement of the nodes isn't really important, but I'll impose that structure here, as it makes it easier to think about what is going on). Place a new node in the middle of each of the original nodes, then join them by edges to form the appropriate triangles. Do this again. And again. And again. In the limit, we end up with the Sierpinski gasket. The first three stages are pictured below:
In the $k$-th stages of this process, we add $3^k$ edges, each of which has length $(frac{1}{2})^k$. Thus the total length of the edges added to the original graph can be computed as
$$ sum_{j=1}^{infty} 3^k left(frac{1}{2}right)^k
= sum_{j=1}^{infty} left(frac{3}{2}right)^k, $$
which is a series diverging to infinity. Since the edges added through this process are dense in the limiting gasket, we can reasonably conclude that the "length" of the Sierpinski gasket is infinite.
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
An alternative construction of the Sierpinski gasket (one that is useful if you want to study differential equations in that setting, for example–see Differential Equations on Fractals by Robert S. Strichartz) is to build it as the limit of a sequence of graphs (a graph is a collection of nodes and edges; you can think of it as a network of cities (nodes) joined by roads (edges)). This is a more direct way of discussing the boundary segments from the original question.
To construct the Sierpinski gasket, start with a graph with three nodes and three edges, arranged to form an equilateral triangle (in graph theory, the arrangement of the nodes isn't really important, but I'll impose that structure here, as it makes it easier to think about what is going on). Place a new node in the middle of each of the original nodes, then join them by edges to form the appropriate triangles. Do this again. And again. And again. In the limit, we end up with the Sierpinski gasket. The first three stages are pictured below:
In the $k$-th stages of this process, we add $3^k$ edges, each of which has length $(frac{1}{2})^k$. Thus the total length of the edges added to the original graph can be computed as
$$ sum_{j=1}^{infty} 3^k left(frac{1}{2}right)^k
= sum_{j=1}^{infty} left(frac{3}{2}right)^k, $$
which is a series diverging to infinity. Since the edges added through this process are dense in the limiting gasket, we can reasonably conclude that the "length" of the Sierpinski gasket is infinite.
An alternative construction of the Sierpinski gasket (one that is useful if you want to study differential equations in that setting, for example–see Differential Equations on Fractals by Robert S. Strichartz) is to build it as the limit of a sequence of graphs (a graph is a collection of nodes and edges; you can think of it as a network of cities (nodes) joined by roads (edges)). This is a more direct way of discussing the boundary segments from the original question.
To construct the Sierpinski gasket, start with a graph with three nodes and three edges, arranged to form an equilateral triangle (in graph theory, the arrangement of the nodes isn't really important, but I'll impose that structure here, as it makes it easier to think about what is going on). Place a new node in the middle of each of the original nodes, then join them by edges to form the appropriate triangles. Do this again. And again. And again. In the limit, we end up with the Sierpinski gasket. The first three stages are pictured below:
In the $k$-th stages of this process, we add $3^k$ edges, each of which has length $(frac{1}{2})^k$. Thus the total length of the edges added to the original graph can be computed as
$$ sum_{j=1}^{infty} 3^k left(frac{1}{2}right)^k
= sum_{j=1}^{infty} left(frac{3}{2}right)^k, $$
which is a series diverging to infinity. Since the edges added through this process are dense in the limiting gasket, we can reasonably conclude that the "length" of the Sierpinski gasket is infinite.
answered Nov 23 at 15:09
Xander Henderson
14k103552
14k103552
add a comment |
add a comment |
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Obviously it diverges (as the sum of a geometric series with ratio $3/2$).
– metamorphy
Nov 23 at 4:00