[math-fun] Geometric probability a la Archimedes
One can prove that the expected distance from a random point on the surface of a sphere to the equatorial plane is half the radius. Assuming we could rephrase this claim in a form that Archimedes would recognize, how would he have proved it? As an example of the kind of proof I would like to see, consider the proposition that the expected distance from a random point in a disk to the boundary of the disk is 1/3 of the radius. One can prove this using the formula for the volume of a cone. (I came up with this myself but I’m sure others have too.) Further examples of the kind of proof I have in mind are Archimedes’ determination of the surface area and volume of the sphere. Jim Propp
It occurs to me that Archimedes, with his love of statics, might have phrased this in terms of the center of mass of a hemispherical shell: it is halfway between the center of the shell and the place where the shell meets its axis of symmetry. Jim Propp On Thu, Jul 25, 2019 at 10:25 AM James Propp <jamespropp@gmail.com> wrote:
One can prove that the expected distance from a random point on the surface of a sphere to the equatorial plane is half the radius. Assuming we could rephrase this claim in a form that Archimedes would recognize, how would he have proved it?
As an example of the kind of proof I would like to see, consider the proposition that the expected distance from a random point in a disk to the boundary of the disk is 1/3 of the radius. One can prove this using the formula for the volume of a cone. (I came up with this myself but I’m sure others have too.)
Further examples of the kind of proof I have in mind are Archimedes’ determination of the surface area and volume of the sphere.
Jim Propp
Jim, Strogatz did quite a bit of research on this for his recent book (pre-calculus ideas). I would first look there. -Veit
On Jul 25, 2019, at 7:25 AM, James Propp <jamespropp@gmail.com> wrote:
One can prove that the expected distance from a random point on the surface of a sphere to the equatorial plane is half the radius. Assuming we could rephrase this claim in a form that Archimedes would recognize, how would he have proved it?
As an example of the kind of proof I would like to see, consider the proposition that the expected distance from a random point in a disk to the boundary of the disk is 1/3 of the radius. One can prove this using the formula for the volume of a cone. (I came up with this myself but I’m sure others have too.)
Further examples of the kind of proof I have in mind are Archimedes’ determination of the surface area and volume of the sphere.
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
The claim follows directly from Archimedes’ “hat-box theorem”. Replace the hemisphere by a cylinder, and apply symmetry! Jim On Thu, Jul 25, 2019 at 10:31 AM Veit Elser <ve10@cornell.edu> wrote:
Jim,
Strogatz did quite a bit of research on this for his recent book (pre-calculus ideas). I would first look there.
-Veit
On Jul 25, 2019, at 7:25 AM, James Propp <jamespropp@gmail.com> wrote:
One can prove that the expected distance from a random point on the surface of a sphere to the equatorial plane is half the radius. Assuming we could rephrase this claim in a form that Archimedes would recognize, how would he have proved it?
As an example of the kind of proof I would like to see, consider the proposition that the expected distance from a random point in a disk to the boundary of the disk is 1/3 of the radius. One can prove this using the formula for the volume of a cone. (I came up with this myself but I’m sure others have too.)
Further examples of the kind of proof I have in mind are Archimedes’ determination of the surface area and volume of the sphere.
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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This reminds me of the lovely fact (mentioned before on this forum) that a random point on a sphere of radius r has a _uniformly_ random z-coordinate from +r to –r. (This is true only in 3 dimensions.) - Cris
On Jul 25, 2019, at 8:30 AM, Veit Elser <ve10@cornell.edu> wrote:
Jim,
Strogatz did quite a bit of research on this for his recent book (pre-calculus ideas). I would first look there.
-Veit
On Jul 25, 2019, at 7:25 AM, James Propp <jamespropp@gmail.com> wrote:
One can prove that the expected distance from a random point on the surface of a sphere to the equatorial plane is half the radius. Assuming we could rephrase this claim in a form that Archimedes would recognize, how would he have proved it?
As an example of the kind of proof I would like to see, consider the proposition that the expected distance from a random point in a disk to the boundary of the disk is 1/3 of the radius. One can prove this using the formula for the volume of a cone. (I came up with this myself but I’m sure others have too.)
Further examples of the kind of proof I have in mind are Archimedes’ determination of the surface area and volume of the sphere.
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Is it really true that the z coordinate is uniformly distributed? A really common question is, "What's the right way to generate pseudo random points on the surface of a sphere?" The most common answer, and the only one I know offhand, is, "Generate three Gaussian coordinate and normalize.". Why have I never heard, "Generate a uniform z coordinate and a uniform azimuth."? On Thu, Jul 25, 2019, 1:47 PM Cris Moore <moore@santafe.edu> wrote:
This reminds me of the lovely fact (mentioned before on this forum) that a random point on a sphere of radius r has a _uniformly_ random z-coordinate from +r to –r. (This is true only in 3 dimensions.)
- Cris
On Jul 25, 2019, at 8:30 AM, Veit Elser <ve10@cornell.edu> wrote:
Jim,
Strogatz did quite a bit of research on this for his recent book (pre-calculus ideas). I would first look there.
-Veit
On Jul 25, 2019, at 7:25 AM, James Propp <jamespropp@gmail.com> wrote:
One can prove that the expected distance from a random point on the surface of a sphere to the equatorial plane is half the radius. Assuming we could rephrase this claim in a form that Archimedes would recognize, how would he have proved it?
As an example of the kind of proof I would like to see, consider the proposition that the expected distance from a random point in a disk to the boundary of the disk is 1/3 of the radius. One can prove this using the formula for the volume of a cone. (I came up with this myself but I’m sure others have too.)
Further examples of the kind of proof I have in mind are Archimedes’ determination of the surface area and volume of the sphere.
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Allan, That's probably because the Archimedes-based method doesn't generalise to S^(n-1) for n != 3. If you take Archimedes's theorem about S^2 and convert it into an equivalent theorem about S^3 (by means of the Hopf fibration), then you get the statement that the squared length of the orthogonal projection of a random unit vector (in R^4) onto a fixed 2-dimensional subspace is uniformly distributed in [0, 1]. Here's an article I wrote recently on the subject: https://cp4space.wordpress.com/2019/05/24/w2-x2-w2-x2-y2-z2/ -- APG.
Sent: Thursday, July 25, 2019 at 7:10 PM From: "Allan Wechsler" <acwacw@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Geometric probability a la Archimedes
Is it really true that the z coordinate is uniformly distributed?
A really common question is, "What's the right way to generate pseudo random points on the surface of a sphere?" The most common answer, and the only one I know offhand, is, "Generate three Gaussian coordinate and normalize.". Why have I never heard, "Generate a uniform z coordinate and a uniform azimuth."?
On Thu, Jul 25, 2019, 1:47 PM Cris Moore <moore@santafe.edu> wrote:
This reminds me of the lovely fact (mentioned before on this forum) that a random point on a sphere of radius r has a _uniformly_ random z-coordinate from +r to –r. (This is true only in 3 dimensions.)
- Cris
On Jul 25, 2019, at 8:30 AM, Veit Elser <ve10@cornell.edu> wrote:
Jim,
Strogatz did quite a bit of research on this for his recent book (pre-calculus ideas). I would first look there.
-Veit
On Jul 25, 2019, at 7:25 AM, James Propp <jamespropp@gmail.com> wrote:
One can prove that the expected distance from a random point on the surface of a sphere to the equatorial plane is half the radius. Assuming we could rephrase this claim in a form that Archimedes would recognize, how would he have proved it?
As an example of the kind of proof I would like to see, consider the proposition that the expected distance from a random point in a disk to the boundary of the disk is 1/3 of the radius. One can prove this using the formula for the volume of a cone. (I came up with this myself but I’m sure others have too.)
Further examples of the kind of proof I have in mind are Archimedes’ determination of the surface area and volume of the sphere.
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Indeed, the following generates uniform points on the unit sphere, where “uniform” means that each bit of area has the right amount of probability: . choose z uniformly in [-1,+1] . choose phi uniformly in [0,2pi] . set x=sqrt(1-z^2) cos phi and y=sqrt(1-z^2) sin phi It’s a very nice exercise. Basically the way the circumference of the circle of latitude decreases as z moves away from zero is exactly compensated for by the way the area of a bit of surface of height dz increases as it tilts from vertical to horizontal. Try it! - Cris
On Jul 25, 2019, at 12:10 PM, Allan Wechsler <acwacw@gmail.com> wrote:
Is it really true that the z coordinate is uniformly distributed?
A really common question is, "What's the right way to generate pseudo random points on the surface of a sphere?" The most common answer, and the only one I know offhand, is, "Generate three Gaussian coordinate and normalize.". Why have I never heard, "Generate a uniform z coordinate and a uniform azimuth."?
On Thu, Jul 25, 2019, 1:47 PM Cris Moore <moore@santafe.edu> wrote:
This reminds me of the lovely fact (mentioned before on this forum) that a random point on a sphere of radius r has a _uniformly_ random z-coordinate from +r to –r. (This is true only in 3 dimensions.)
- Cris
On Jul 25, 2019, at 8:30 AM, Veit Elser <ve10@cornell.edu> wrote:
Jim,
Strogatz did quite a bit of research on this for his recent book (pre-calculus ideas). I would first look there.
-Veit
On Jul 25, 2019, at 7:25 AM, James Propp <jamespropp@gmail.com> wrote:
One can prove that the expected distance from a random point on the surface of a sphere to the equatorial plane is half the radius. Assuming we could rephrase this claim in a form that Archimedes would recognize, how would he have proved it?
As an example of the kind of proof I would like to see, consider the proposition that the expected distance from a random point in a disk to the boundary of the disk is 1/3 of the radius. One can prove this using the formula for the volume of a cone. (I came up with this myself but I’m sure others have too.)
Further examples of the kind of proof I have in mind are Archimedes’ determination of the surface area and volume of the sphere.
Jim Propp _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Cris Moore moore@santafe.edu The disorders and miseries which result gradually incline the minds of men to seek security and repose in the absolute power of an individual; and sooner or later the chief of some prevailing faction, more able or more fortunate than his competitors, turns this disposition to the purposes of his own elevation, on the ruins of public liberty. — George Washington
This method relies on the theorem that slices of a sphere of equal thickness have equal area. But this algorithm has a problem that sqrt(1-z^2) has large large relative error when z is near 1 or -1. If I were generating random points on the sphere, I'd use the Gaussian algorithm, and it would work in any dimension. -- Gene On Thursday, July 25, 2019, 12:00:42 PM PDT, Cris Moore <moore@santafe.edu> wrote: Indeed, the following generates uniform points on the unit sphere, where “uniform” means that each bit of area has the right amount of probability: . choose z uniformly in [-1,+1] . choose phi uniformly in [0,2pi] . set x=sqrt(1-z^2) cos phi and y=sqrt(1-z^2) sin phi It’s a very nice exercise. Basically the way the circumference of the circle of latitude decreases as z moves away from zero is exactly compensated for by the way the area of a bit of surface of height dz increases as it tilts from vertical to horizontal. Try it! - Cris
On Jul 25, 2019, at 12:10 PM, Allan Wechsler <acwacw@gmail.com> wrote:
Is it really true that the z coordinate is uniformly distributed?
A really common question is, "What's the right way to generate pseudo random points on the surface of a sphere?" The most common answer, and the only one I know offhand, is, "Generate three Gaussian coordinate and normalize.". Why have I never heard, "Generate a uniform z coordinate and a uniform azimuth."?
On Thu, Jul 25, 2019, 1:47 PM Cris Moore <moore@santafe.edu> wrote:
This reminds me of the lovely fact (mentioned before on this forum) that a random point on a sphere of radius r has a _uniformly_ random z-coordinate from +r to –r. (This is true only in 3 dimensions.)
- Cris
On Jul 25, 2019, at 8:30 AM, Veit Elser <ve10@cornell.edu> wrote:
Jim,
Strogatz did quite a bit of research on this for his recent book (pre-calculus ideas). I would first look there.
-Veit
On Jul 25, 2019, at 7:25 AM, James Propp <jamespropp@gmail.com> wrote:
One can prove that the expected distance from a random point on the surface of a sphere to the equatorial plane is half the radius. Assuming we could rephrase this claim in a form that Archimedes would recognize, how would he have proved it?
As an example of the kind of proof I would like to see, consider the proposition that the expected distance from a random point in a disk to the boundary of the disk is 1/3 of the radius. One can prove this using the formula for the volume of a cone. (I came up with this myself but I’m sure others have too.)
Further examples of the kind of proof I have in mind are Archimedes’ determination of the surface area and volume of the sphere.
Jim Propp
I guess Gene's explanation is probably the main reason that the Gaussian algorithm is the one that gets used, mostly. It's too bad that generating Gaussian samples is so clunky. Sqrt(-2 ln U) cis V, indeed. On Thu, Jul 25, 2019 at 3:11 PM Eugene Salamin via math-fun < math-fun@mailman.xmission.com> wrote:
This method relies on the theorem that slices of a sphere of equal thickness have equal area. But this algorithm has a problem that sqrt(1-z^2) has large large relative error when z is near 1 or -1. If I were generating random points on the sphere, I'd use the Gaussian algorithm, and it would work in any dimension.
-- Gene
On Thursday, July 25, 2019, 12:00:42 PM PDT, Cris Moore < moore@santafe.edu> wrote:
Indeed, the following generates uniform points on the unit sphere, where “uniform” means that each bit of area has the right amount of probability:
. choose z uniformly in [-1,+1] . choose phi uniformly in [0,2pi] . set x=sqrt(1-z^2) cos phi and y=sqrt(1-z^2) sin phi
It’s a very nice exercise. Basically the way the circumference of the circle of latitude decreases as z moves away from zero is exactly compensated for by the way the area of a bit of surface of height dz increases as it tilts from vertical to horizontal. Try it!
- Cris
On Jul 25, 2019, at 12:10 PM, Allan Wechsler <acwacw@gmail.com> wrote:
Is it really true that the z coordinate is uniformly distributed?
A really common question is, "What's the right way to generate pseudo random points on the surface of a sphere?" The most common answer, and the only one I know offhand, is, "Generate three Gaussian coordinate and normalize.". Why have I never heard, "Generate a uniform z coordinate and a uniform azimuth."?
On Thu, Jul 25, 2019, 1:47 PM Cris Moore <moore@santafe.edu> wrote:
This reminds me of the lovely fact (mentioned before on this forum)
that a
random point on a sphere of radius r has a _uniformly_ random z-coordinate from +r to –r. (This is true only in 3 dimensions.)
- Cris
On Jul 25, 2019, at 8:30 AM, Veit Elser <ve10@cornell.edu> wrote:
Jim,
Strogatz did quite a bit of research on this for his recent book (pre-calculus ideas). I would first look there.
-Veit
On Jul 25, 2019, at 7:25 AM, James Propp <jamespropp@gmail.com> wrote:
One can prove that the expected distance from a random point on the surface of a sphere to the equatorial plane is half the radius. Assuming we could rephrase this claim in a form that Archimedes would recognize, how would he have proved it?
As an example of the kind of proof I would like to see, consider the proposition that the expected distance from a random point in a disk to the boundary of the disk is 1/3 of the radius. One can prove this using the formula for the volume of a cone. (I came up with this myself but I’m sure others have too.)
Further examples of the kind of proof I have in mind are Archimedes’ determination of the surface area and volume of the sphere.
Jim Propp
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participants (6)
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Allan Wechsler -
Cris Moore -
Eugene Salamin -
James Propp -
Veit Elser