Series for $pi$ which correspond to apollonian gaskets or hyperbolic tilings of the unit disk
$begingroup$
Consider the two partitions of the unit disk in $mathbb{R}^{2}$, the first an Apollonian gasket and the second is the ${7,3}$ hyperbolic tiling:
Since the unit disk has radius $1$, both of these have area $pi$, and for each partition, there is a series representation of $pi$, so:
- Given an Apollonian gasket in the disk of unit radius, is there an straightforward way of getting the sequence of areas of disks which comprise it?
- Given a regular hyperbolic tiling, is there a straightfoward way of obtaining the Euclidean area of the (distorted) polygons which comprise it?
geometry circles area hyperbolic-geometry pi
edited Jan 16 at 7:41
Glorfindel
3,41581830
asked May 24 '13 at 21:16
graveolensagraveolensa
3,09711738
$endgroup$
add a comment |
$begingroup$
Consider the two partitions of the unit disk in $mathbb{R}^{2}$, the first an Apollonian gasket and the second is the ${7,3}$ hyperbolic tiling:
Since the unit disk has radius $1$, both of these have area $pi$, and for each partition, there is a series representation of $pi$, so:
- Given an Apollonian gasket in the disk of unit radius, is there an straightforward way of getting the sequence of areas of disks which comprise it?
- Given a regular hyperbolic tiling, is there a straightfoward way of obtaining the Euclidean area of the (distorted) polygons which comprise it?
geometry circles area hyperbolic-geometry pi
edited Jan 16 at 7:41
Glorfindel
3,41581830
asked May 24 '13 at 21:16
graveolensagraveolensa
3,09711738
$endgroup$
$begingroup$
The first one would give a series with pi in it because it's made of circles the 2nd seems more interesting. Still I like this idea for making series for pi visual. Kudos! Are those heptagons similar? or related by stretching?
$endgroup$
– futurebird
May 25 '13 at 13:53
add a comment |
$begingroup$
Consider the two partitions of the unit disk in $mathbb{R}^{2}$, the first an Apollonian gasket and the second is the ${7,3}$ hyperbolic tiling:
Since the unit disk has radius $1$, both of these have area $pi$, and for each partition, there is a series representation of $pi$, so:
- Given an Apollonian gasket in the disk of unit radius, is there an straightforward way of getting the sequence of areas of disks which comprise it?
- Given a regular hyperbolic tiling, is there a straightfoward way of obtaining the Euclidean area of the (distorted) polygons which comprise it?
geometry circles area hyperbolic-geometry pi
edited Jan 16 at 7:41
Glorfindel
3,41581830
asked May 24 '13 at 21:16
graveolensagraveolensa
3,09711738
$endgroup$
Consider the two partitions of the unit disk in $mathbb{R}^{2}$, the first an Apollonian gasket and the second is the ${7,3}$ hyperbolic tiling:
Since the unit disk has radius $1$, both of these have area $pi$, and for each partition, there is a series representation of $pi$, so:
- Given an Apollonian gasket in the disk of unit radius, is there an straightforward way of getting the sequence of areas of disks which comprise it?
- Given a regular hyperbolic tiling, is there a straightfoward way of obtaining the Euclidean area of the (distorted) polygons which comprise it?
geometry circles area hyperbolic-geometry pi
geometry circles area hyperbolic-geometry pi
edited Jan 16 at 7:41
Glorfindel
3,41581830
asked May 24 '13 at 21:16
graveolensagraveolensa
3,09711738
edited Jan 16 at 7:41
Glorfindel
3,41581830
asked May 24 '13 at 21:16
graveolensagraveolensa
3,09711738
edited Jan 16 at 7:41
Glorfindel
3,41581830
edited Jan 16 at 7:41
Glorfindel
3,41581830
edited Jan 16 at 7:41
Glorfindel
3,41581830
3,41581830
asked May 24 '13 at 21:16
graveolensagraveolensa
3,09711738
asked May 24 '13 at 21:16
graveolensagraveolensa
3,09711738
asked May 24 '13 at 21:16
graveolensagraveolensa
3,09711738
3,09711738
$begingroup$
The first one would give a series with pi in it because it's made of circles the 2nd seems more interesting. Still I like this idea for making series for pi visual. Kudos! Are those heptagons similar? or related by stretching?
$endgroup$
– futurebird
May 25 '13 at 13:53
add a comment |
$begingroup$
The first one would give a series with pi in it because it's made of circles the 2nd seems more interesting. Still I like this idea for making series for pi visual. Kudos! Are those heptagons similar? or related by stretching?
$endgroup$
– futurebird
May 25 '13 at 13:53
$begingroup$
The first one would give a series with pi in it because it's made of circles the 2nd seems more interesting. Still I like this idea for making series for pi visual. Kudos! Are those heptagons similar? or related by stretching?
$endgroup$
– futurebird
May 25 '13 at 13:53
$begingroup$
The first one would give a series with pi in it because it's made of circles the 2nd seems more interesting. Still I like this idea for making series for pi visual. Kudos! Are those heptagons similar? or related by stretching?
$endgroup$
– futurebird
May 25 '13 at 13:53
add a comment |
1 Answer
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$begingroup$
Although you could find such sequences, they would most likely themselves make heavy use of $pi$. For this reason, the best you could hope for would be some sequence from which you could factor out $pi$ such that the result adds up to $1$.
Apollonian gasket
You can use Descartes' theorem to compute the (inverse) circle radii, starting from four given ones. For the symmetric configuration your picture suggests, these will be non-integral (see Wikipedia on almost $D_3$ symmetry in this context). More precisely, according to this section of Erich's packing center the radii of the thee largest circles will be $r=frac{3}{2sqrt3 + 3}$. So you can compute subsequent radii from these, given enough patience. To turn all these radii back into areas you'd square them and multiply them by $pi$. So if you learn that the squared radii sum up to $1$, this does not tell you one bit about the value of $pi$.
Hyperbolic tiling
It seems you're using the Poincaré disk model of hyperbolic geometry. Which means your heptagon boundaries are not really straight lines but instead circle arcs, which makes the Euclidean area computation both painfully complicated to write and very intimately related to circles. Therefore I doubt you can get the area expressed without massive use of $pi$. So the whole result would most likely appear like a circular argument, pardon the pun. One alternative might be using hyperbolic corners but connecting them with Euclidean straight lines. Another alternative might be the Beltrami-Klein model. Neither does look particularly suited for an easy sequence description of $pi$, though.
answered May 27 '13 at 11:36
MvGMvG
31.1k450106
$endgroup$
add a comment |
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$begingroup$
Although you could find such sequences, they would most likely themselves make heavy use of $pi$. For this reason, the best you could hope for would be some sequence from which you could factor out $pi$ such that the result adds up to $1$.
Apollonian gasket
You can use Descartes' theorem to compute the (inverse) circle radii, starting from four given ones. For the symmetric configuration your picture suggests, these will be non-integral (see Wikipedia on almost $D_3$ symmetry in this context). More precisely, according to this section of Erich's packing center the radii of the thee largest circles will be $r=frac{3}{2sqrt3 + 3}$. So you can compute subsequent radii from these, given enough patience. To turn all these radii back into areas you'd square them and multiply them by $pi$. So if you learn that the squared radii sum up to $1$, this does not tell you one bit about the value of $pi$.
Hyperbolic tiling
It seems you're using the Poincaré disk model of hyperbolic geometry. Which means your heptagon boundaries are not really straight lines but instead circle arcs, which makes the Euclidean area computation both painfully complicated to write and very intimately related to circles. Therefore I doubt you can get the area expressed without massive use of $pi$. So the whole result would most likely appear like a circular argument, pardon the pun. One alternative might be using hyperbolic corners but connecting them with Euclidean straight lines. Another alternative might be the Beltrami-Klein model. Neither does look particularly suited for an easy sequence description of $pi$, though.
answered May 27 '13 at 11:36
MvGMvG
31.1k450106
$endgroup$
add a comment |
$begingroup$
Although you could find such sequences, they would most likely themselves make heavy use of $pi$. For this reason, the best you could hope for would be some sequence from which you could factor out $pi$ such that the result adds up to $1$.
Apollonian gasket
You can use Descartes' theorem to compute the (inverse) circle radii, starting from four given ones. For the symmetric configuration your picture suggests, these will be non-integral (see Wikipedia on almost $D_3$ symmetry in this context). More precisely, according to this section of Erich's packing center the radii of the thee largest circles will be $r=frac{3}{2sqrt3 + 3}$. So you can compute subsequent radii from these, given enough patience. To turn all these radii back into areas you'd square them and multiply them by $pi$. So if you learn that the squared radii sum up to $1$, this does not tell you one bit about the value of $pi$.
Hyperbolic tiling
It seems you're using the Poincaré disk model of hyperbolic geometry. Which means your heptagon boundaries are not really straight lines but instead circle arcs, which makes the Euclidean area computation both painfully complicated to write and very intimately related to circles. Therefore I doubt you can get the area expressed without massive use of $pi$. So the whole result would most likely appear like a circular argument, pardon the pun. One alternative might be using hyperbolic corners but connecting them with Euclidean straight lines. Another alternative might be the Beltrami-Klein model. Neither does look particularly suited for an easy sequence description of $pi$, though.
answered May 27 '13 at 11:36
MvGMvG
31.1k450106
$endgroup$
add a comment |
$begingroup$
Although you could find such sequences, they would most likely themselves make heavy use of $pi$. For this reason, the best you could hope for would be some sequence from which you could factor out $pi$ such that the result adds up to $1$.
Apollonian gasket
You can use Descartes' theorem to compute the (inverse) circle radii, starting from four given ones. For the symmetric configuration your picture suggests, these will be non-integral (see Wikipedia on almost $D_3$ symmetry in this context). More precisely, according to this section of Erich's packing center the radii of the thee largest circles will be $r=frac{3}{2sqrt3 + 3}$. So you can compute subsequent radii from these, given enough patience. To turn all these radii back into areas you'd square them and multiply them by $pi$. So if you learn that the squared radii sum up to $1$, this does not tell you one bit about the value of $pi$.
Hyperbolic tiling
It seems you're using the Poincaré disk model of hyperbolic geometry. Which means your heptagon boundaries are not really straight lines but instead circle arcs, which makes the Euclidean area computation both painfully complicated to write and very intimately related to circles. Therefore I doubt you can get the area expressed without massive use of $pi$. So the whole result would most likely appear like a circular argument, pardon the pun. One alternative might be using hyperbolic corners but connecting them with Euclidean straight lines. Another alternative might be the Beltrami-Klein model. Neither does look particularly suited for an easy sequence description of $pi$, though.
answered May 27 '13 at 11:36
MvGMvG
31.1k450106
$endgroup$
Although you could find such sequences, they would most likely themselves make heavy use of $pi$. For this reason, the best you could hope for would be some sequence from which you could factor out $pi$ such that the result adds up to $1$.
Apollonian gasket
You can use Descartes' theorem to compute the (inverse) circle radii, starting from four given ones. For the symmetric configuration your picture suggests, these will be non-integral (see Wikipedia on almost $D_3$ symmetry in this context). More precisely, according to this section of Erich's packing center the radii of the thee largest circles will be $r=frac{3}{2sqrt3 + 3}$. So you can compute subsequent radii from these, given enough patience. To turn all these radii back into areas you'd square them and multiply them by $pi$. So if you learn that the squared radii sum up to $1$, this does not tell you one bit about the value of $pi$.
Hyperbolic tiling
It seems you're using the Poincaré disk model of hyperbolic geometry. Which means your heptagon boundaries are not really straight lines but instead circle arcs, which makes the Euclidean area computation both painfully complicated to write and very intimately related to circles. Therefore I doubt you can get the area expressed without massive use of $pi$. So the whole result would most likely appear like a circular argument, pardon the pun. One alternative might be using hyperbolic corners but connecting them with Euclidean straight lines. Another alternative might be the Beltrami-Klein model. Neither does look particularly suited for an easy sequence description of $pi$, though.
answered May 27 '13 at 11:36
MvGMvG
31.1k450106
answered May 27 '13 at 11:36
MvGMvG
31.1k450106
answered May 27 '13 at 11:36
MvGMvG
31.1k450106
answered May 27 '13 at 11:36
MvGMvG
31.1k450106
31.1k450106
add a comment |
add a comment |
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$begingroup$
The first one would give a series with pi in it because it's made of circles the 2nd seems more interesting. Still I like this idea for making series for pi visual. Kudos! Are those heptagons similar? or related by stretching?
$endgroup$
– futurebird
May 25 '13 at 13:53