FPGA Central

A sphere is a three dimensional figure enclosed by the locus of all
points that are of distance r (called the radius) away from the center
point. It has a volume of 4/3 pi r^3 and a surface area of 4 pi r^2.

A circle is the two-dimensional analog of a sphere. It is a two
dimensional figure enclosed by the locus of all points that are a
distance of r away from the center point. It has an area of pi r^2 and a
circumference of 2 pi r.

A line is a one-dimensional analog of a sphere (OK, I'm reaching, but
bear with me). It is a one dimensional figure enclosed by the locus of
all (two) points that are a distance of r away from the center point. It
has a length of 2 r and a -- uh -- oh never mind.

Now hopefully I have some momentum: The four dimensional analog of a
sphere (commonly called a hypersphere) is a four dimensional figure
enclosed by the locus of all points that are a distance of r away from
the center point. It contains an amount of space (not a volume,
certainly -- hypervolume?) equal to 1/2 pi^2 r^4, and has a 'surface'
volume of 2 pi^2 r^3.

I've done the math in a way that seems obvious to me (integrate the
volume of the sphere that forms the surface of the hypersphere as one of
the dimensions varies from -r to r), and that pi^2 just _belongs_ there,
no matter how much it surprises me.

So, am I right? Wrong? Anyone done this calculation before?

Thanks. (and no, this isn't idle speculation; I actually need to know to
solve a DSP problem I'm wrestling with).

--
http://www.wescottdesign.com

Tim Wescott <tim@seemywebsite.com> wrote:
> A sphere is a three dimensional figure enclosed by the locus of all
> points that are of distance r (called the radius) away from the center
> point. It has a volume of 4/3 pi r^3 and a surface area of 4 pi r^2.
(snip)

> Now hopefully I have some momentum: The four dimensional analog of a
> sphere (commonly called a hypersphere) is a four dimensional figure
> enclosed by the locus of all points that are a distance of r away from
> the center point. It contains an amount of space (not a volume,
> certainly -- hypervolume?) equal to 1/2 pi^2 r^4, and has a 'surface'
> volume of 2 pi^2 r^3.

The surface volume agrees with mathworld. The hypervolume seems
to be an easy integral, so I agree with that one, too.

-- glen

>Now hopefully I have some momentum: The four dimensional analog of a
>sphere (commonly called a hypersphere) is a four dimensional figure
>enclosed by the locus of all points that are a distance of r away from
>the center point. It contains an amount of space (not a volume,
>certainly -- hypervolume?) equal to 1/2 pi^2 r^4, and has a 'surface'
>volume of 2 pi^2 r^3.
>
>I've done the math in a way that seems obvious to me (integrate the
>volume of the sphere that forms the surface of the hypersphere as one o

>the dimensions varies from -r to r), and that pi^2 just _belongs_ there

>no matter how much it surprises me.


Both the formulae for S.A. & volume for a 4D sphere are correct but if u
logic of integration is followed then the pi squared term cant be accounte
for (if u integrate wrt only r ) .
Generally the hypervolume & the hyper surface area of a
dimensional sphere( of which urs is a N=4 case) is not calculated by th
method mentioned by u but the result rather arrives thru principle o
mathematical induction & the result involves beta & gamma functions whic
is why u get to see the powers of pi (e.g. gamma(1/2) = sqrt(pi) and th
general formula involves powers of gamma(1/2))
Remarkably for a n-sphere as it is generally called , u get hype
surface area = derivative of hypervolume , again theory of beta n gamm
functions is involved in this , so once u know the volume of a hyperspher
, u differentiate it wrt r to get hypersurface area.
The formula for a n-sphere hypervolume is
Vn(r) = (gamma(1/2))^n * r^n/ gamma(1/2*n +1)
&
Sn(r) = d/dr(Vn(r))

Note : The above discussion applies to N dimensional Euclidea
space.

Regards
Yogesh P. Gharote
Bangalore,
India

On Jun 20, 4:00*am, "yogesh_gharote" <yogesh_ghar...@yahoo.com> wrote:
<<<material snipped>>

> * * * * *Remarkably for a n-sphere as it is generally called , u get hyper
> surface area = derivative of hypervolume , again theory of beta n gamma
> functions is involved in this , so once u know the volume of a hypersphere
> , u differentiate it wrt r to get hypersurface area.
> * * * * * The formula for a n-sphere hypervolume is
> * * * * *Vn(r) = (gamma(1/2))^n * r^n/ gamma(1/2*n +1)
> &
> * * * * *Sn(r) = d/dr(Vn(r))


The volume of a sphere in n-dimensional space is
proportional to r^n, and, as Yogesh states, we have
to use the theory of beta and gamma functions to
find the value of the constant of proportionality. But
the basic result asked for by Tim and stated as

Sn(r) = d/dr(Vn(r))

above should be intuitively obvious even if one does
not know the value of the constant of proportionality.
Consider the thin shell of thickness dr formed by
scooping out the sphere of radius r from the inside of
of a sphere of radius r+dr. The volume of this shell
is Vn(r+dr) - Vn(r). But since the shell has surface
area approximately Sn(r) and thickness dr, its volume
is approximately Sn(r)dr. Equate the two expressions,
divide both sides by dr, take limits as dr goes to 0.
Finicky epsilon-delta folks are advised to use the
approach once espoused by Julius Kusuma in
this newsgroup ("The most rigorous arguments shall
be proved by the most vigorous hand waving") as
needed to establish the result.

Incidentally, I don't understand what Tim meant when
he wrote

>integrate the volume of the sphere that forms the
>surface of the hypersphere as one of the dimensions
>varies from -r to r

The antiderivative of 2\pi r is \pi r^2, the antiderivative
of 4\pi r^2 is (4/3)\pi r^3, and so on and so forth, and
there is no need to let "one of the dimensions" vary
from -r to + r.

Hope this helps

--Dilip Sarwate

On Jun 20, 7:40*am, "dvsarw...@yahoo.com" <dvsarw...@gmail.com> wrote:
[snip]
> Finicky epsilon-delta folks are advised to use the
> approach once espoused by Julius Kusuma in
> this newsgroup ("The most rigorous arguments shall
> be proved by the most vigorous hand waving") as
> needed to establish the result.
>
[snip]

Dilip, this is the most flattering citation I ever received! :-)

Julius

On Sat, 20 Jun 2009 04:00:47 -0500, yogesh_gharote wrote:

>>Now hopefully I have some momentum: The four dimensional analog of a
>>sphere (commonly called a hypersphere) is a four dimensional figure
>>enclosed by the locus of all points that are a distance of r away from
>>the center point. It contains an amount of space (not a volume,
>>certainly -- hypervolume?) equal to 1/2 pi^2 r^4, and has a 'surface'
>>volume of 2 pi^2 r^3.
>>
>>I've done the math in a way that seems obvious to me (integrate the
>>volume of the sphere that forms the surface of the hypersphere as one of
>
>>the dimensions varies from -r to r), and that pi^2 just _belongs_ there,
>
>>no matter how much it surprises me.
>
>
> Both the formulae for S.A. & volume for a 4D sphere are correct but if
> ur logic of integration is followed then the pi squared term cant be
> accounted for (if u integrate wrt only r ) .
> Generally the hypervolume & the hyper surface area of a N
> dimensional sphere( of which urs is a N=4 case) is not calculated by the
> method mentioned by u but the result rather arrives thru principle of
> mathematical induction & the result involves beta & gamma functions
> which is why u get to see the powers of pi (e.g. gamma(1/2) = sqrt(pi)
> and the general formula involves powers of gamma(1/2))
> Remarkably for a n-sphere as it is generally called , u get
> hyper
> surface area = derivative of hypervolume , again theory of beta n gamma
> functions is involved in this , so once u know the volume of a
> hypersphere , u differentiate it wrt r to get hypersurface area.
> The formula for a n-sphere hypervolume is
> Vn(r) = (gamma(1/2))^n * r^n/ gamma(1/2*n +1)
> &
> Sn(r) = d/dr(Vn(r))
>
> Note : The above discussion applies to N dimensional Euclidean
> space.

My choice of oddball integrals was intentional, as I want to go on to
calculating various moments for the probability distributions of the
surface of the hypersphere when 3D probability distributions are mapped
onto it. Clearly if I map a tight Gaussian distribution onto the
hypersphere with a standard deviation that's much smaller than the
hypersphere radius the resulting probability distribution will be easy;
it's figuring out what happens as that probability distribution opens up
that's making my brain cramp.


--
http://www.wescottdesign.com

On 20 Jun, 19:50, Tim Wescott <t...@seemywebsite.com> wrote:

> My choice of oddball integrals was intentional, as I want to go on to
> calculating various moments for the probability distributions of the
> surface of the hypersphere when 3D probability distributions are mapped
> onto it. *Clearly if I map a tight Gaussian distribution onto the
> hypersphere with a standard deviation that's much smaller than the
> hypersphere radius the resulting probability distribution will be easy;
> it's figuring out what happens as that probability distribution opens up
> that's making my brain cramp.

I'm a bit curious about what kind of problem leads you
out in such kinds of calculations?

Rune

On Sat, 20 Jun 2009 11:04:38 -0700, Rune Allnor wrote:

> On 20 Jun, 19:50, Tim Wescott <t...@seemywebsite.com> wrote:
>
>> My choice of oddball integrals was intentional, as I want to go on to
>> calculating various moments for the probability distributions of the
>> surface of the hypersphere when 3D probability distributions are mapped
>> onto it. Â*Clearly if I map a tight Gaussian distribution onto the
>> hypersphere with a standard deviation that's much smaller than the
>> hypersphere radius the resulting probability distribution will be easy;
>> it's figuring out what happens as that probability distribution opens
>> up that's making my brain cramp.
>
> I'm a bit curious about what kind of problem leads you out in such kinds
> of calculations?

Unscented transformations for quaternion PDFs used to represent angles in
a (hopefully soon-to-be) unscented Kalman filter. The math needed to
represent body rotations is quite hairy; to date unit-length quaternions
seems to be the best approach, although getting this PDF stuff figured
out is proving to be interesting, at best.

--
http://www.wescottdesign.com

On 20 Jun, 20:14, Tim Wescott <t...@seemywebsite.com> wrote:
> On Sat, 20 Jun 2009 11:04:38 -0700, Rune Allnor wrote:
> > On 20 Jun, 19:50, Tim Wescott <t...@seemywebsite.com> wrote:
>
> >> My choice of oddball integrals was intentional, as I want to go on to
> >> calculating various moments for the probability distributions of the
> >> surface of the hypersphere when 3D probability distributions are mapped
> >> onto it. *Clearly if I map a tight Gaussian distribution onto the
> >> hypersphere with a standard deviation that's much smaller than the
> >> hypersphere radius the resulting probability distribution will be easy;
> >> it's figuring out what happens as that probability distribution opens
> >> up that's making my brain cramp.
>
> > I'm a bit curious about what kind of problem leads you out in such kinds
> > of calculations?
>
> Unscented transformations for quaternion PDFs used to represent angles in
> a (hopefully soon-to-be) unscented Kalman filter.

The expected value is a 4-vector pointing in some desired
direction and the PDF represents the probability distribution
of actual directions?

Rune

On Sat, 20 Jun 2009 11:37:27 -0700, Rune Allnor wrote:

> On 20 Jun, 20:14, Tim Wescott <t...@seemywebsite.com> wrote:
>> On Sat, 20 Jun 2009 11:04:38 -0700, Rune Allnor wrote:
>> > On 20 Jun, 19:50, Tim Wescott <t...@seemywebsite.com> wrote:
>>
>> >> My choice of oddball integrals was intentional, as I want to go on
>> >> to calculating various moments for the probability distributions of
>> >> the surface of the hypersphere when 3D probability distributions are
>> >> mapped onto it. Â*Clearly if I map a tight Gaussian distribution onto
>> >> the hypersphere with a standard deviation that's much smaller than
>> >> the hypersphere radius the resulting probability distribution will
>> >> be easy; it's figuring out what happens as that probability
>> >> distribution opens up that's making my brain cramp.
>>
>> > I'm a bit curious about what kind of problem leads you out in such
>> > kinds of calculations?
>>
>> Unscented transformations for quaternion PDFs used to represent angles
>> in a (hopefully soon-to-be) unscented Kalman filter.
>
> The expected value is a 4-vector pointing in some desired direction and
> the PDF represents the probability distribution of actual directions?

Yup. Given a PDF I'd _like_ to be able to solve for the expected value
of the four vector elements, as well as their cross-correlation. I'd
like this to be in a form that's tractable enough that I can code it into
an algorithm without either making people's heads explode when they read
it and without making the processor bog down.

But for now I'll just settle with being able to get the element means and
variances out of the thing for a variety of variances of the Gaussians,
or a clear indication that if I insist on using Gaussians the math is
going to be hopelessly intractable.

--
http://www.wescottdesign.com

Tim Wescott wrote:

> Yup. Given a PDF I'd _like_ to be able to solve for the
> expected value of the four vector elements, as well as their
> cross-correlation. I'd like this to be in a form that's
> tractable enough that I can code it into an algorithm without
> either making people's heads explode when they read it and
> without making the processor bog down.
>
> But for now I'll just settle with being able to get the element
> means and variances out of the thing for a variety of variances
> of the Gaussians, or a clear indication that if I insist on
> using Gaussians the math is going to be hopelessly intractable.

How have you decided to project the Gaussian from Euclidean space to
the 3-sphere?


Martin

--
Quidquid latine scriptum est, altum videtur.

On Sun, 21 Jun 2009 02:23:11 +0000, Martin Eisenberg wrote:

> Tim Wescott wrote:
>
>> Yup. Given a PDF I'd _like_ to be able to solve for the expected value
>> of the four vector elements, as well as their cross-correlation. I'd
>> like this to be in a form that's tractable enough that I can code it
>> into an algorithm without either making people's heads explode when
>> they read it and without making the processor bog down.
>>
>> But for now I'll just settle with being able to get the element means
>> and variances out of the thing for a variety of variances of the
>> Gaussians, or a clear indication that if I insist on using Gaussians
>> the math is going to be hopelessly intractable.
>
> How have you decided to project the Gaussian from Euclidean space to the
> 3-sphere?
>
>
> Martin

You mean from Euclidean 3-space to the surface of the 4-sphere? I'm
still trying to figure that out!

I'm finding this an interesting challenge -- I'm not nearly as strong in
this sort of mathematics as I am with visualization. Unfortunately, I
have yet to succeed at visualizing four dimensions (silly me). So I'm
getting better at the math as fast as I can.

--
http://www.wescottdesign.com

On 20 Jun, 23:44, Tim Wescott <t...@seemywebsite.com> wrote:
> On Sat, 20 Jun 2009 11:37:27 -0700, Rune Allnor wrote:
> > On 20 Jun, 20:14, Tim Wescott <t...@seemywebsite.com> wrote:
> >> On Sat, 20 Jun 2009 11:04:38 -0700, Rune Allnor wrote:
> >> > On 20 Jun, 19:50, Tim Wescott <t...@seemywebsite.com> wrote:
>
> >> >> My choice of oddball integrals was intentional, as I want to go on
> >> >> to calculating various moments for the probability distributions of
> >> >> the surface of the hypersphere when 3D probability distributions are
> >> >> mapped onto it. *Clearly if I map a tight Gaussian distribution onto
> >> >> the hypersphere with a standard deviation that's much smaller than
> >> >> the hypersphere radius the resulting probability distribution will
> >> >> be easy; it's figuring out what happens as that probability
> >> >> distribution opens up that's making my brain cramp.
>
> >> > I'm a bit curious about what kind of problem leads you out in such
> >> > kinds of calculations?
>
> >> Unscented transformations for quaternion PDFs used to represent angles
> >> in a (hopefully soon-to-be) unscented Kalman filter.
>
> > The expected value is a 4-vector pointing in some desired direction and
> > the PDF represents the probability distribution of actual directions?
>
> Yup. *Given a PDF I'd _like_ to be able to solve for the expected value
> of the four vector elements, as well as their cross-correlation. *I'd
> like this to be in a form that's tractable enough that I can code it into
> an algorithm without either making people's heads explode when they read
> it and without making the processor bog down.
>
> But for now I'll just settle with being able to get the element means and
> variances out of the thing for a variety of variances of the Gaussians,
> or a clear indication that if I insist on using Gaussians the math is
> going to be hopelessly intractable.

I did have a go at this for a Gaussian on the unit circle .
Note that I did this at 4AM (no intoxication though), arithmetics
has never been a force of mine, my pencil broke and I generally
am not able to read my own handwriting.

In other words, don't trust the details in what follows:

================================================== ===========
Leaving out the scaling factors, the Gaussian PDF on the
circle becomes (view with fixed-width font)

inf
g(phi) = sum exp-((phi - nu)/sigma + 2 n pi)^2 [1]
n = -inf

since there is a possibility that the angle error in reality
includes any number of revolutions around the unit circle.

For simplicity, define

a = (phi - nu)/sigma [2a]
b = 2pi [2b]

and substitute into [1]:

inf
g(phi) = sum exp-(a + bn)^2 [3]
n = -inf

inf
= sum exp-(a^2+2abn+b^2n^2) [4]
n = -inf

inf
= exp(a^2) sum exp(-2abn)exp(-b^2 n^2) [5]
n = -inf

Next, convert from a double-sided infinite sum to a
one-sided infinite sum, using

inf inf
sum exp(-x) = 1 + 2 sum cosh(x). [6]
n = -inf n=1

Apply [6] to the first exponential inside the summation
in [5] to find

inf
g(phi) = exp(a^2){1 + 2 sum cosh(2abn)exp(-b^2 n^2)} [7]
n=1

To get the end result, substitute [2a-b] into [7] and insert
the missing scaling factors.
================================================== ===========

That's as far as I can get. Because of the n^2 fector in
the exponent inside the sum, the summation formula for
geometric series can't be used (at least that's how I
understand it).

Now, the form of [7] seems to be rather nice: It's a basic
Gaussian (the outer term) and a lot of correction terms.
Since the exponential correction term is dominated by n^2,
you probably don't need to many correction terms to get an
impression of the PDF. Note also that the cosh term contains
the (phi-nu)/sigma correction, so let sigma -> inf and
investigate what hapepns when the PDF 'opens up' towards
uniform. Maybe one is able to work this out further,
using power series etc.

Or you can mage a worst-case assumption about the series
being a sum of cosh(2abn)exp(-b^2 n) terms (substituted
n for n^2 in the exponent) and use the formula for a
geometric series to proceed.

Now, you do introduce an error, but you push the PDF
towards something 'worse' in the sense

g(phi)_n > g(phi)_n^2,

where the subscript indicates what n term you used in
the exponential.

Whatever blunders and errors I might have made in the
above, here is the maths challenge:

Now, this excercise (or the corrected / amended version
of it...) works well when working on a unit circle in
2D space. Even generalizing it to the spheric shell in
3D space is beyond me: The above works for the azimuth
angle in the spherical coordinate system. I have no idea
how to handle the elevation angle: It covers only half
the sphere (from 'pole' to 'pole') and once it extends
beyond +/- pi, it couples back in on the azimuth angle,
which flips by pi.

I don't see how to handle this for the problem at hand.

But, as somebody made a point of earlier this week:
That's what mathematicians are for - handle the
technicalities in the theory we mere mortal engineers
have to use.

Any takers among the mathematicians? How does one
extend the 'infintely many wrap-arounds' on the
circle in 2D space to a sphere in 3D space?

Rune

Tim Wescott wrote:
> On Sun, 21 Jun 2009 02:23:11 +0000, Martin Eisenberg wrote:

>> How have you decided to project the Gaussian from Euclidean
>> space to the 3-sphere?

> You mean from Euclidean 3-space to the surface of the 4-sphere?

I was using the mathematical terminology where the n-sphere is the
shell with intrinsic dimension n, so the set of unit quaternions is a
3-sphere in this lingo.

You can visualize it just like I gather you did your content
integrals: Imagine that a tiny ball appears in midair and grows. It
decelerates, reaches a maximum radius, and collapses back into a
point. What you saw is a 4-ball falling through E^3, with the time
segment isomorphic to a diameter and each (3-ball, instant) pair a
slice through the 4D solid. Likewise, the 3-sphere is the disjoint
union of (2-sphere, instant) pairs, and you can identify the time
segment with the quaternion rotation angle and each sphere with the
set of axis directions.

> I'm still trying to figure that out!

My earlier Riemann sphere idea generalizes to stereographic
projection, but the Von Mises-Fisher distribution might be more
useful in terms of available theory.


Martin

--
Quidquid latine scriptum est, altum videtur.

On Jun 21, 11:35*am, Rune Allnor <all...@tele.ntnu.no> wrote:
> On 20 Jun, 23:44, Tim Wescott <t...@seemywebsite.com> wrote:
>
>
>
>
>
> > On Sat, 20 Jun 2009 11:37:27 -0700, Rune Allnor wrote:
> > > On 20 Jun, 20:14, Tim Wescott <t...@seemywebsite.com> wrote:
> > >> On Sat, 20 Jun 2009 11:04:38 -0700, Rune Allnor wrote:
> > >> > On 20 Jun, 19:50, Tim Wescott <t...@seemywebsite.com> wrote:
>
> > >> >> My choice of oddball integrals was intentional, as I want to go on
> > >> >> to calculating various moments for the probability distributions of
> > >> >> the surface of the hypersphere when 3D probability distributions are
> > >> >> mapped onto it. *Clearly if I map a tight Gaussian distributiononto
> > >> >> the hypersphere with a standard deviation that's much smaller than
> > >> >> the hypersphere radius the resulting probability distribution will
> > >> >> be easy; it's figuring out what happens as that probability
> > >> >> distribution opens up that's making my brain cramp.
>
> > >> > I'm a bit curious about what kind of problem leads you out in such
> > >> > kinds of calculations?
>
> > >> Unscented transformations for quaternion PDFs used to represent angles
> > >> in a (hopefully soon-to-be) unscented Kalman filter.
>
> > > The expected value is a 4-vector pointing in some desired direction and
> > > the PDF represents the probability distribution of actual directions?
>
> > Yup. *Given a PDF I'd _like_ to be able to solve for the expected value
> > of the four vector elements, as well as their cross-correlation. *I'd
> > like this to be in a form that's tractable enough that I can code it into
> > an algorithm without either making people's heads explode when they read
> > it and without making the processor bog down.
>
> > But for now I'll just settle with being able to get the element means and
> > variances out of the thing for a variety of variances of the Gaussians,
> > or a clear indication that if I insist on using Gaussians the math is
> > going to be hopelessly intractable.
>
> I did have a go at this for a Gaussian on the unit circle .
> Note that I did this at 4AM (no intoxication though), arithmetics
> has never been a force of mine, my pencil broke and I generally
> am not able to read my own handwriting.
>
> In other words, don't trust the details in what follows:
>
> ================================================== ===========
> Leaving out the scaling factors, the Gaussian PDF on the
> circle becomes (view with fixed-width font)
>
> * * * * * *inf
> g(phi) = * sum * * exp-((phi - nu)/sigma + 2 n pi)^2 *[1]
> * * * * *n = -inf
>
> since there is a possibility that the angle error in reality
> includes any number of revolutions around the unit circle.
>
> For simplicity, define
>
> a = (phi - nu)/sigma * * * * * * * * * * * * * * * * *[2a]
> b = 2pi * * * * * * * * * * * * * * * ** * * * * * * [2b]
>
> and substitute into [1]:
>
> * * * * * *inf
> g(phi) = * sum * *exp-(a + bn)^2 * * * * * * * * * * *[3]
> * * * * *n = -inf
>
> * * * * * *inf
> * * * *= * sum * *exp-(a^2+2abn+b^2n^2) * * * * * * * [4]
> * * * * *n = -inf
>
> * * * * * * * * * * *inf
> * * * *= *exp(a^2) * sum * * exp(-2abn)exp(-b^2 n^2) *[5]
> * * * * * * * * * *n = -inf
>
> Next, convert from a double-sided infinite sum to a
> one-sided infinite sum, using
>
> * inf * * * * * * * * * * * *inf
> * sum * * exp(-x) * *= 1 + 2 sum cosh(x). * * * * ** [6]
> n = -inf * * * * * * * * * * n=1
>
> Apply [6] to the first exponential inside the summation
> in [5] to find
>
> * * * * * * * * * * * * inf
> g(phi) = exp(a^2){1 + 2 sum cosh(2abn)exp(-b^2 n^2)} *[7]
> * * * * * * * * * * * * n=1
>
> To get the end result, substitute [2a-b] into [7] and insert
> the missing scaling factors.
> ================================================== ===========
>
> That's as far as I can get. Because of the n^2 fector in
> the exponent inside the sum, the summation formula for
> geometric series can't be used (at least that's how I
> understand it).
>
> Now, the form of [7] seems to be rather nice: It's a basic
> Gaussian (the outer term) and a lot of correction terms.
> Since the exponential correction term is dominated by n^2,
> you probably don't need to many correction terms to get an
> impression of the PDF. Note also that the cosh term contains
> the (phi-nu)/sigma correction, so let sigma -> inf and
> investigate what hapepns when the PDF 'opens up' towards
> uniform. Maybe one is able to work this out further,
> using power series etc.
>
> Or you can mage a worst-case assumption about the series
> being a sum of cosh(2abn)exp(-b^2 n) terms (substituted
> n for n^2 in the exponent) and use the formula for a
> geometric series to proceed.
>
> Now, you do introduce an error, but you push the PDF
> towards something 'worse' in the sense
>
> * * * * * *g(phi)_n > *g(phi)_n^2,
>
> where the subscript indicates what n term you used in
> the exponential.
>
> Whatever blunders and errors I might have made in the
> above, here is the maths challenge:
>
> Now, this excercise (or the corrected / amended version
> of it...) works well when working on a unit circle in
> 2D space. Even generalizing it to the spheric shell in
> 3D space is beyond me: The above works for the azimuth
> angle in the spherical coordinate system. I have no idea
> how to handle the elevation angle: It covers only half
> the sphere (from 'pole' to 'pole') and once it extends
> beyond +/- pi, it couples back in on the azimuth angle,
> which flips by pi.
>
> I don't see how to handle this for the problem at hand.
>
> But, as somebody made a point of earlier this week:
> That's what mathematicians are for - handle the
> technicalities in the theory we mere mortal engineers
> have to use.
>
> Any takers among the mathematicians? How does one
> extend the 'infintely many wrap-arounds' on the
> circle in 2D space to a sphere in 3D space?
>
> Rune

You extend it via the geodesics that start at the mean and wrap round
and round the sphere, using distance along the geodesic as a
coordinate. This corresponds to identifying circular sets of points in
R^{n}, where n is the dimensionality of the sphere (embedded in R^{n +
1}), just as the circle case (n = 1) corresponds to identifying all
the odd integers and all the even integers. But one has to be careful
about the underlying measure also. This has to be the measure derived
from the constant curvature metric on the sphere, i.e. that induced by
its embedding in Euclidean space.

But the Gaussian used in this way doesn't make much sense, by which I
mean there is no compelling argument for using it. There are other
distributions that reduce to a Gaussian for small variances (i.e. when
one can approximate the situation by the tangent space at the mean),
e.g. the von Mises distribution, and that are better defined on a
periodic domain. The general area is known as directional statistics.

illywhacker;

On 21 Jun, 16:04, illywhacker <illywac...@gmail.com> wrote:
> On Jun 21, 11:35*am, Rune Allnor <all...@tele.ntnu.no> wrote:
>
>
>
>
>
> > On 20 Jun, 23:44, Tim Wescott <t...@seemywebsite.com> wrote:
>
> > > On Sat, 20 Jun 2009 11:37:27 -0700, Rune Allnor wrote:
> > > > On 20 Jun, 20:14, Tim Wescott <t...@seemywebsite.com> wrote:
> > > >> On Sat, 20 Jun 2009 11:04:38 -0700, Rune Allnor wrote:
> > > >> > On 20 Jun, 19:50, Tim Wescott <t...@seemywebsite.com> wrote:
>
> > > >> >> My choice of oddball integrals was intentional, as I want to goon
> > > >> >> to calculating various moments for the probability distributions of
> > > >> >> the surface of the hypersphere when 3D probability distributions are
> > > >> >> mapped onto it. *Clearly if I map a tight Gaussian distribution onto
> > > >> >> the hypersphere with a standard deviation that's much smaller than
> > > >> >> the hypersphere radius the resulting probability distribution will
> > > >> >> be easy; it's figuring out what happens as that probability
> > > >> >> distribution opens up that's making my brain cramp.
>
> > > >> > I'm a bit curious about what kind of problem leads you out in such
> > > >> > kinds of calculations?
>
> > > >> Unscented transformations for quaternion PDFs used to represent angles
> > > >> in a (hopefully soon-to-be) unscented Kalman filter.
>
> > > > The expected value is a 4-vector pointing in some desired directionand
> > > > the PDF represents the probability distribution of actual directions?
>
> > > Yup. *Given a PDF I'd _like_ to be able to solve for the expected value
> > > of the four vector elements, as well as their cross-correlation. *I'd
> > > like this to be in a form that's tractable enough that I can code it into
> > > an algorithm without either making people's heads explode when they read
> > > it and without making the processor bog down.
>
> > > But for now I'll just settle with being able to get the element meansand
> > > variances out of the thing for a variety of variances of the Gaussians,
> > > or a clear indication that if I insist on using Gaussians the math is
> > > going to be hopelessly intractable.
>
> > I did have a go at this for a Gaussian on the unit circle .
> > Note that I did this at 4AM (no intoxication though), arithmetics
> > has never been a force of mine, my pencil broke and I generally
> > am not able to read my own handwriting.
>
> > In other words, don't trust the details in what follows:
>
> > ================================================== ===========
> > Leaving out the scaling factors, the Gaussian PDF on the
> > circle becomes (view with fixed-width font)
>
> > * * * * * *inf
> > g(phi) = * sum * * exp-((phi - nu)/sigma + 2 n pi)^2 *[1]
> > * * * * *n = -inf
>
> > since there is a possibility that the angle error in reality
> > includes any number of revolutions around the unit circle.
>
> > For simplicity, define
>
> > a = (phi - nu)/sigma * * * * * * * * * * * * * * * * *[2a]
> > b = 2pi * * * * * * * * * * * * * * * * * * * * * * * [2b]
>
> > and substitute into [1]:
>
> > * * * * * *inf
> > g(phi) = * sum * *exp-(a + bn)^2 * * * * * * * * * * *[3]
> > * * * * *n = -inf
>
> > * * * * * *inf
> > * * * *= * sum * *exp-(a^2+2abn+b^2n^2) * * * ** * * [4]
> > * * * * *n = -inf
>
> > * * * * * * * * * * *inf
> > * * * *= *exp(a^2) * sum * * exp(-2abn)exp(-b^2 n^2) *[5]
> > * * * * * * * * * *n = -inf
>
> > Next, convert from a double-sided infinite sum to a
> > one-sided infinite sum, using
>
> > * inf * * * * * * * * * * * *inf
> > * sum * * exp(-x) * *= 1 + 2 sum cosh(x). * * * * * * [6]
> > n = -inf * * * * * * * * * * n=1
>
> > Apply [6] to the first exponential inside the summation
> > in [5] to find
>
> > * * * * * * * * * * * * inf
> > g(phi) = exp(a^2){1 + 2 sum cosh(2abn)exp(-b^2 n^2)} *[7]
> > * * * * * * * * * * * * n=1
>
> > To get the end result, substitute [2a-b] into [7] and insert
> > the missing scaling factors.
> > ================================================== ===========
>
> > That's as far as I can get. Because of the n^2 fector in
> > the exponent inside the sum, the summation formula for
> > geometric series can't be used (at least that's how I
> > understand it).
>
> > Now, the form of [7] seems to be rather nice: It's a basic
> > Gaussian (the outer term) and a lot of correction terms.
> > Since the exponential correction term is dominated by n^2,
> > you probably don't need to many correction terms to get an
> > impression of the PDF. Note also that the cosh term contains
> > the (phi-nu)/sigma correction, so let sigma -> inf and
> > investigate what hapepns when the PDF 'opens up' towards
> > uniform. Maybe one is able to work this out further,
> > using power series etc.
>
> > Or you can mage a worst-case assumption about the series
> > being a sum of cosh(2abn)exp(-b^2 n) terms (substituted
> > n for n^2 in the exponent) and use the formula for a
> > geometric series to proceed.
>
> > Now, you do introduce an error, but you push the PDF
> > towards something 'worse' in the sense
>
> > * * * * * *g(phi)_n > *g(phi)_n^2,
>
> > where the subscript indicates what n term you used in
> > the exponential.
>
> > Whatever blunders and errors I might have made in the
> > above, here is the maths challenge:
>
> > Now, this excercise (or the corrected / amended version
> > of it...) works well when working on a unit circle in
> > 2D space. Even generalizing it to the spheric shell in
> > 3D space is beyond me: The above works for the azimuth
> > angle in the spherical coordinate system. I have no idea
> > how to handle the elevation angle: It covers only half
> > the sphere (from 'pole' to 'pole') and once it extends
> > beyond +/- pi, it couples back in on the azimuth angle,
> > which flips by pi.
>
> > I don't see how to handle this for the problem at hand.
>
> > But, as somebody made a point of earlier this week:
> > That's what mathematicians are for - handle the
> > technicalities in the theory we mere mortal engineers
> > have to use.
>
> > Any takers among the mathematicians? How does one
> > extend the 'infintely many wrap-arounds' on the
> > circle in 2D space to a sphere in 3D space?
>
> > Rune
>
> You extend it via the geodesics that start at the mean and wrap round
> and round the sphere, using distance along the geodesic as a
> coordinate. This corresponds to identifying circular sets of points in
> R^{n}, where n is the dimensionality of the sphere (embedded in R^{n +
> 1}), just as the circle case (n = 1) corresponds to identifying all
> the odd integers and all the even integers. But one has to be careful
> about the underlying measure also. This has to be the measure derived
> from the constant curvature metric on the sphere, i.e. that induced by
> its embedding in Euclidean space.
>
> But the Gaussian used in this way doesn't make much sense, by which I
> mean there is no compelling argument for using it. There are other
> distributions that reduce to a Gaussian for small variances (i.e. when
> one can approximate the situation by the tangent space at the mean),
> e.g. the von Mises distribution, and that are better defined on a
> periodic domain. The general area is known as directional statistics.

MAybe I misunderstand Tim's problem, but in an earlier thread,

http://groups.google.no/group/comp.d...1254af9?hl=no#

he said something about variance becoming large and the
distribution approaching uniform.

Rune

On Jun 20, 6:27*am, Tim Wescott <t...@seemywebsite.com> wrote:
> A sphere is a three dimensional figure enclosed by the locus of all
> points that are of distance r (called the radius) away from the center
> point. *It has a volume of 4/3 pi r^3 and a surface area of 4 pi r^2.
>
> A circle is the two-dimensional analog of a sphere. *It is a two
> dimensional figure enclosed by the locus of all points that are a
> distance of r away from the center point. *It has an area of pi r^2 anda
> circumference of 2 pi r.
>
> A line is a one-dimensional analog of a sphere (OK, I'm reaching, but
> bear with me). *It is a one dimensional figure enclosed by the locus of
> all (two) points that are a distance of r away from the center point. *It
> has a length of 2 r and a -- uh -- oh never mind.
>
> Now hopefully I have some momentum: *The four dimensional analog of a
> sphere (commonly called a hypersphere) is a four dimensional figure
> enclosed by the locus of all points that are a distance of r away from
> the center point. *It contains an amount of space (not a volume,
> certainly -- hypervolume?) equal to 1/2 pi^2 r^4, and has a 'surface'
> volume of 2 pi^2 r^3.
>
> I've done the math in a way that seems obvious to me (integrate the
> volume of the sphere that forms the surface of the hypersphere as one of
> the dimensions varies from -r to r), and that pi^2 just _belongs_ there,
> no matter how much it surprises me.
>
> So, am I right? *Wrong? *Anyone done this calculation before?
>
> Thanks. *(and no, this isn't idle speculation; I actually need to know to
> solve a DSP problem I'm wrestling with).

'Directional statistics'.

illywhacker;

On Mon, 22 Jun 2009 01:43:28 -0700, illywhacker wrote:

On Jun 20, 6:27Â*am, Tim Wescott <t...@seemywebsite.com> wrote:
> A sphere is a three dimensional figure enclosed by the locus of all
> points that are of distance r (called the radius) away from the center
> point. Â*It has a volume of 4/3 pi r^3 and a surface area of 4 pi r^2.
>
> A circle is the two-dimensional analog of a sphere. Â*It is a two
> dimensional figure enclosed by the locus of all points that are a
> distance of r away from the center point. Â*It has an area of pi r^2 and
a
> circumference of 2 pi r.
>
> A line is a one-dimensional analog of a sphere (OK, I'm reaching, but
> bear with me). Â*It is a one dimensional figure enclosed by the locus of
> all (two) points that are a distance of r away from the center point.
Â*It
> has a length of 2 r and a -- uh -- oh never mind.
>
> Now hopefully I have some momentum: Â*The four dimensional analog of a
> sphere (commonly called a hypersphere) is a four dimensional figure
> enclosed by the locus of all points that are a distance of r away from
> the center point. Â*It contains an amount of space (not a volume,
> certainly -- hypervolume?) equal to 1/2 pi^2 r^4, and has a 'surface'
> volume of 2 pi^2 r^3.
>
> I've done the math in a way that seems obvious to me (integrate the
> volume of the sphere that forms the surface of the hypersphere as one of
> the dimensions varies from -r to r), and that pi^2 just _belongs_ there,
> no matter how much it surprises me.
>
> So, am I right? Â*Wrong? Â*Anyone done this calculation before?
>
> Thanks. Â*(and no, this isn't idle speculation; I actually need to know
to
> solve a DSP problem I'm wrestling with).
:
:'Directional statistics'.
:
:illywhacker;

A short web search shows that this is a book length subject, and one that
may require more than one book to get ready to read the book on the
subject.

I'll probably have to muddle through as best as I can with this (I'm
starting to think that some sort of uniform PDF on a 3-D plane, mapped to
the hypersphere surface, may get me sufficient performance without making
my brain explode).

So -- got any book recommendations? I've got statistics, estimation and
detection theory, a bit of functional analysis, and Kalman filtering
under my belt, but it's mostly from an engineering perspective rather
than a 'math wonk' perspective. It would be nice if I could get just one
book that does a good job of teaching the run-in (what the hell is a
Riemann Manifold?), rather than having to go buy books just so I can read
the book.

--
www.wescottdesign.com

On Jun 22, 10:33*am, Tim Wescott <t...@seemywebsite.com> wrote:
> On Mon, 22 Jun 2009 01:43:28 -0700, illywhacker wrote:
>
> On Jun 20, 6:27*am, Tim Wescott <t...@seemywebsite.com> wrote:
>
>
>
> > A sphere is a three dimensional figure enclosed by the locus of all
> > points that are of distance r (called the radius) away from the center
> > point. *It has a volume of 4/3 pi r^3 and a surface area of 4 pi r^2.
>
> > A circle is the two-dimensional analog of a sphere. *It is a two
> > dimensional figure enclosed by the locus of all points that are a
> > distance of r away from the center point. *It has an area of pi r^2 and
> a
> > circumference of 2 pi r.
>
> > A line is a one-dimensional analog of a sphere (OK, I'm reaching, but
> > bear with me). *It is a one dimensional figure enclosed by the locus of
> > all (two) points that are a distance of r away from the center point.
> *It
> > has a length of 2 r and a -- uh -- oh never mind.
>
> > Now hopefully I have some momentum: *The four dimensional analog of a
> > sphere (commonly called a hypersphere) is a four dimensional figure
> > enclosed by the locus of all points that are a distance of r away from
> > the center point. *It contains an amount of space (not a volume,
> > certainly -- hypervolume?) equal to 1/2 pi^2 r^4, and has a 'surface'
> > volume of 2 pi^2 r^3.
>
> > I've done the math in a way that seems obvious to me (integrate the
> > volume of the sphere that forms the surface of the hypersphere as one of
> > the dimensions varies from -r to r), and that pi^2 just _belongs_ there,
> > no matter how much it surprises me.
>
> > So, am I right? *Wrong? *Anyone done this calculation before?
>
> > Thanks. *(and no, this isn't idle speculation; I actually need to know
> to
> > solve a DSP problem I'm wrestling with).
>
> :
> :'Directional statistics'.
> :
> :illywhacker;
>
> A short web search shows that this is a book length subject, and one that
> may require more than one book to get ready to read the book on the
> subject.
>
> I'll probably have to muddle through as best as I can with this (I'm
> starting to think that some sort of uniform PDF on a 3-D plane, mapped to
> the hypersphere surface, may get me sufficient performance without making
> my brain explode).
>
> So -- got any book recommendations? *I've got statistics, estimation and
> detection theory, a bit of functional analysis, and Kalman filtering
> under my belt, but it's mostly from an engineering perspective rather
> than a 'math wonk' perspective. *It would be nice if I could get just one
> book that does a good job of teaching the run-in (what the hell is a
> Riemann Manifold?), rather than having to go buy books just so I can read
> the book.
>
> --www.wescottdesign.com- Hide quoted text -
>
> - Show quoted text -

Tim,
I have no idea if this will help, but you might try looking at a
couple of things on quaternion calculus.

http://www.geometrictools.com/Docume...uaternions.pdf

http://www.euclideanspace.com/math/d...ulus/index.htm

These might get you to the point where you can read the book ;>)

Maurice Givens

"Tim Wescott" <tim@seemywebsite.com> wrote in message
news:v8udnYrLb_C1-KHXnZ2dnUVZ_jCdnZ2d@web-ster.com...
>A sphere is a three dimensional figure enclosed by the locus of all
> points that are of distance r (called the radius) away from the center
> point. It has a volume of 4/3 pi r^3 and a surface area of 4 pi r^2.
>
> A circle is the two-dimensional analog of a sphere. It is a two
> dimensional figure enclosed by the locus of all points that are a
> distance of r away from the center point. It has an area of pi r^2 and a
> circumference of 2 pi r.
>
> A line is a one-dimensional analog of a sphere (OK, I'm reaching, but
> bear with me). It is a one dimensional figure enclosed by the locus of
> all (two) points that are a distance of r away from the center point. It
> has a length of 2 r and a -- uh -- oh never mind.
>
> Now hopefully I have some momentum: The four dimensional analog of a
> sphere (commonly called a hypersphere) is a four dimensional figure
> enclosed by the locus of all points that are a distance of r away from
> the center point. It contains an amount of space (not a volume,
> certainly -- hypervolume?) equal to 1/2 pi^2 r^4, and has a 'surface'
> volume of 2 pi^2 r^3.
>
> I've done the math in a way that seems obvious to me (integrate the
> volume of the sphere that forms the surface of the hypersphere as one of
> the dimensions varies from -r to r), and that pi^2 just _belongs_ there,
> no matter how much it surprises me.
>
> So, am I right? Wrong? Anyone done this calculation before?
>
> Thanks. (and no, this isn't idle speculation; I actually need to know to
> solve a DSP problem I'm wrestling with).
>
> --
> http://www.wescottdesign.com

I worked out the area and volume of an n-sphere many years ago. The method
I had of expressing the results was not as elegant as that in Wikipedia
http://en.wikipedia.org/wiki/Hypersphere but the formula you got is correct.

Best wishes,
--Phil Martel

>On Sat, 20 Jun 2009 11:04:38 -0700, Rune Allnor wrote:
>
>> On 20 Jun, 19:50, Tim Wescott <t...@seemywebsite.com> wrote:
>>
>>> My choice of oddball integrals was intentional, as I want to go on to
>>> calculating various moments for the probability distributions of the
>>> surface of the hypersphere when 3D probability distributions ar
mapped
>>> onto it. Â*Clearly if I map a tight Gaussian distribution onto the
>>> hypersphere with a standard deviation that's much smaller than the
>>> hypersphere radius the resulting probability distribution will b
easy;
>>> it's figuring out what happens as that probability distribution opens
>>> up that's making my brain cramp.
>>
>> I'm a bit curious about what kind of problem leads you out in suc
kinds
>> of calculations?
>
>Unscented transformations for quaternion PDFs used to represent angles i

>a (hopefully soon-to-be) unscented Kalman filter. The math needed to
>represent body rotations is quite hairy; to date unit-length quaternion

>seems to be the best approach, although getting this PDF stuff figured
>out is proving to be interesting, at best.
>

Search for the paper "Unscented Filtering in a Unit Quaternion Space fo
Spacecraft Attitude Estimation" ...it's on IEEE. This problem has bee
attacked before, and that paper seemed to be a good read. I have not trie
implementing it, but I could probably trudge through it eventually.

On Tue, 23 Jun 2009 20:45:54 -0500, Michael Plante wrote:

>>On Sat, 20 Jun 2009 11:04:38 -0700, Rune Allnor wrote:
>>
>>> On 20 Jun, 19:50, Tim Wescott <t...@seemywebsite.com> wrote:
>>>
>>>> My choice of oddball integrals was intentional, as I want to go on to
>>>> calculating various moments for the probability distributions of the
>>>> surface of the hypersphere when 3D probability distributions are
> mapped
>>>> onto it. Â*Clearly if I map a tight Gaussian distribution onto the
>>>> hypersphere with a standard deviation that's much smaller than the
>>>> hypersphere radius the resulting probability distribution will be
> easy;
>>>> it's figuring out what happens as that probability distribution opens
>>>> up that's making my brain cramp.
>>>
>>> I'm a bit curious about what kind of problem leads you out in such
> kinds
>>> of calculations?
>>
>>Unscented transformations for quaternion PDFs used to represent angles
>>in
>
>>a (hopefully soon-to-be) unscented Kalman filter. The math needed to
>>represent body rotations is quite hairy; to date unit-length quaternions
>
>>seems to be the best approach, although getting this PDF stuff figured
>>out is proving to be interesting, at best.
>>
>>
> Search for the paper "Unscented Filtering in a Unit Quaternion Space for
> Spacecraft Attitude Estimation" ...it's on IEEE. This problem has been
> attacked before, and that paper seemed to be a good read. I have not
> tried implementing it, but I could probably trudge through it
> eventually.

Well, except for the "spacecraft" part that could just about be the title
of what I'm doing.

That may even be worthwhile to pay for.

--
www.wescottdesign.com