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server science forum beginner
Joined: 24 Mar 2005
Posts: 26

Posted: Thu Mar 24, 2005 9:51 pm Post subject:
How real are the "Virtual" partticles?



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Dave L. Renfro science forum Guru
Joined: 29 Apr 2005
Posts: 570

Posted: Thu Mar 24, 2005 9:51 pm Post subject:
Re: fractional iteration of functions



qmagick@yahoo.com
[sci.math.research: January 28, 2005 14:00:05 +0000 (UTC)]
http://mathforum.org/epigone/sci.math.research/dralyoysli
wrote (in part):
Quote:  Wow, first, thanks for all the responses. I now have more
then enough references to investigate. I have gotten a good
response on this question. Second, I would like to respond
to Mr. Geisler's last comment about axiomatic basis for
function iteration. I think that will be the goal of the
paper I write. Well, at least an axiomatic basis for well
behaved functions over the complex plane.

Here's another reference that you might want to look at.
(I didn't see it among those suggested in this thread.)
Daniel S. Alexander, "A History of Complex Dynamics from
Schröder to Fatou and Julia", Aspects of Mathematics E 24,
Friedr. Vieweg & Sohn, Braunschweig, 1994.
[MR 95d:01014; Zbl 788.30001]
http://www.emis.de/cgibin/MATHitem?0788.30001
The Zbl review is especially long (the URL above takes you
to a publically available webpage). Two additional reviews
that I know of are:
Theodore W. Gamelin, Historia Mathematica 23 (1996), 7484
Robert B. Burckel, SIAM Review 36(4) (Dec. 1994), 663664.
Alexander's book is useful for its survey of early work
on what you're interested in. For example, Section 2.2
"Analytic Iteration", is preceded by this paragraph:
"Before reviewing the responses of Korkine and Farkas to
Schröder's study of functional equations it will be useful
to first say a few words about analytic iteration, and
then to briefly outline the respective approaches of
Schröder, Korkine and Farkas to this problem." (p. 24)
Dave L. Renfro 

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korifey science forum beginner
Joined: 24 Mar 2005
Posts: 1

Posted: Thu Mar 24, 2005 9:51 pm Post subject:
Re: Set Family on a Graph



Aldar Chan ÐÉÛÅÔ:
Given a graph G=(V, E) where V and E are the sets of vertices
and edges respectively. Also given a set P which could be a set
of secret keys. I want to assign a subset of P to each vertex of
G in such a way that given any two vertices v1 and v2 (both in
V), the assigned subsets for them, denoted by p1 and p2
respectively, would have a nonempty intersection which is not
covered by the union of the assigned subsets of a bounded
number of any other vertices. What is the minimum size of P
needed for that and what is the maximum size of the subset
assigned to a node? Is there an efficient algorithm which can
do that? Any related literature appeared before?
ÒÏÌ 

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mdshafri science forum beginner
Joined: 24 Mar 2005
Posts: 1

Posted: Thu Mar 24, 2005 9:51 pm Post subject:
Re: Hellinger Distance



How to set the range of d value to represent the measurement.
We could be set 0 for no changes between two discrete distributions C
and H. What happen if the value of C far much bigger than H.
Please anybody advice me !!
On 19 Apr 1997 13:46:35 0500, Herman Rubin wrote:
Quote:  In article <335237AD.59E2@dcs.rhbnc.ac.uk>,
Peter Burge <peteb@dcs.rhbnc.ac.uk> wrote:
Please could someone give me a reference for a measure
known as the Hellinger Distance between two
discrete distributions C and H. In Latex,
d=\sum_{i=0}^{K} (\sqrt{C_{i}}  \sqrt{H_{i}})^{2}
This expression may be lacking additional terms.
It is not lacking any terms. For general measures, although
it is not likely to be of much use unless they are finite,
it is
d = \int (sqrt(dF)  sqrt(dG))^2.
That this is welldefined can be seen by using as a base
measure H = F + G. It is the supremum of the discrete
versions obtained by using finite partitions.
The real introduction of this into mathematics was by
Kakutani, who gives credit to Hellinger in a footnote.

This address is for information only. I do not claim that these
views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Dept. of Statistics, Purdue Univ., West Lafayette
IN479071399
hrubin@stat.purdue.edu Phone: (765)4946054 FAX:
(765)4940558 


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Annales de Toulouse science forum beginner
Joined: 13 Jun 2005
Posts: 3

Posted: Thu Mar 24, 2005 9:51 pm Post subject:
Annales de Toulouse, 1/2005



The issue 1/2005 of the Annales de la faculte des sciences de Toulouse
has appeared.
Contents:
Michel Hickel
Sur quelques aspects de la geometrie de l'espace des arcs traces sur un
espace analytique
Laurent Bernis
Solutions stationnaires des equations de VlasovPoisson a symetries
cylindriques
Duc Tai Trinh
Coefficients de Stokes du modele cubique : point de vue de la
resurgence quantique
Mohammad Daher
Translations mesurables et ensembles de Rosenthal
Andrzej J. Maciejewski, Maria Przybylska
Differential Galois approach to the nonintegrability of the heavy top
problem
The abstracts and some full texts can be downloaded at:
http://picard.upstlse.fr/~annales (french version)
http://picard.upstlse.fr/~annales/index_en.html (english version) 

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Urs Schreiber science forum Guru Wannabe
Joined: 04 May 2005
Posts: 127

Posted: Thu Mar 24, 2005 9:51 pm Post subject:
fibered categories vs. 2sheaves



Ettore Aldrovandi <ealdrov@zeno.math.fsu.edu> wrote in message news:<20050202181810.GC24685@math.fsu.edu>...
Quote:  In article <ctbmgs$b8s$1@news.ks.uiuc.edu>,
Urs Schreiber <Urs.Schreiber@uniessen.de> wrote:
"John Baez" <baez@math.removethis.ucr.andthis.edu> schrieb im Newsbeitrag
news:ctas57$1p2$1@news.ks.uiuc.edu...
2bundles are meant to be an alternative to gerbes: although
I've done my best to hide it above, a gerbe is really more
like a categorified *sheaf* than a bundle. And, just as a
bundle has a sheaf of sections, we're hoping that a 2bundle
has a stack of sections, which in certain cases will be a
gerbe. That's one of the things we need to figure out,
though.
First, by definition, a gerbe is a stack is a fibered category. The concept
"fibered category" is a categorification of the concept "presheaf". But not
its full categorification. I am wondering if the full categorification of
the concept "presheaf" has been studied before, and under which name.
More precisely, what I am talking about is this:
A presheaf over a topological space X is a morphism in Cat,
namely a (contravariant) functor from the category O(X) of open
subsets of X to Set.
Hi, you are maybe rigidifying the starting situation a bit too
much? A gerbe is a locally nonempty and locally connected stack
in groupoids. In particular, it is a fibered category. Now, this
does not make it a contravariant functor in Cat,

Yes. Please note that I did not claim that it is a functor in Cat!
I said that a *presheaf* is a contravariant functor and hence a
morphism in Cat. Then I said that a gerbe is almost but not quite the
*categorification* of this.
There is a principle "categorification by internalization" that
suggests that the categorification of a mathematical concept which is
a morphism in Cat should be a morphism in 2Cat, the 3category of
2categories (some flavor of it, at least).
Therefore I said that the categorification of "presheaf" should be
something like a 2functor from the 2category O(S) of open
sub2spaces to Cat, the 2category of categories. This is what I
wanted to call a "2presheaf".
There is an issue here with how to precisely define O(S), but in
general this definition does reproduce the freedom of having natural
transformations of the kind that you are referring to:
Quote:  if p: G > X is a fibered category over X, say G is a gerbe, but
id doesn't matter here, if a, b, c are objects of X, then you
have the corresponding fiber categories G(a), G(b), G(c).If
i:b>a
is a morphism in X, then there is a corresponding "restriction"
functor
i^*: G(a) > G(b).
and if now j: c > b is another morphism with the corresponding
functor
j^*: G(b) > G(c)
then there is only a natural transformation
(ij)^* ==> j^*i^*
between the two resulting functors from G(a) to G(c).

Yes, the same holds true for the 2sheaves that I tentatively talked
about. Here there are 2morphisms in O(S), namely natural
transformations between "inclusion"functors and these are taken to
2morphisms in Cat, namely natural transformations between functors
between the G(a), G(b), G(c).
My point was that a fibered category is much like a 2presheaf but
with the 2morphisms in the source ignored. But thanks for your remark
about "lax functors". Maybe my question is clarified by noting how a
fibered category is to be thought of not as the categorification of a
presheaf, but as some sort of "laxification".
If you think I don't make sense please let me know! 

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Herman Rubin science forum Guru
Joined: 25 Mar 2005
Posts: 730

Posted: Thu Mar 24, 2005 9:51 pm Post subject:
Re: de Finetti's theorem



In article <4xk44kwf6yd0@legacy>, Ruitao Zhang <rzhang@jhsph.edu> wrote:
Quote:  de Finetti theorem said that if an infinite sequence is exchangeable
then there exist a unique probability measure such that the de finetti
hold. Is this probability measure the prior measure. In Bayesian we
choose start from any prior measure. Here it seems that the prior is
unique. I don't understand it.

The de Finetti theorem is a probability theorem, not a
statistical method. In probability, the measure is
unique. The mixing measure is the tail measure.
For example, if one has a Polya urn scheme with b black
balls and w white balls, the distribution of draws will
be the same as if the balls drawn were independent
Bernoulli trials with the probability of black drawn
from a Beta(b,w) distribution.
Quote:  For infinite population with exchangeability, we can use relative
frequency of an event to estimate the the probability of this event.
But, For the finite population, why we need exchangeability?

For finite populations, exchangeability does not even
give as much. The trials given the whole finite
population are not independent, as the "draws" are
without replacement.

This address is for information only. I do not claim that these views
are those of the Statistics Department or of Purdue University.
Herman Rubin, Department of Statistics, Purdue University
hrubin@stat.purdue.edu Phone: (765)4946054 FAX: (765)4940558 

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W. Dale Hall science forum Guru
Joined: 29 Apr 2005
Posts: 350

Posted: Thu Mar 24, 2005 9:51 pm Post subject:
Re: orientability of the universal bundle



philippe wrote:
Quote:  Does anyone know for which $n$ the manifold {(x,D) in Rn * G1(Rn)
such that x in D } is orientable AS A MANIFOLD (and not as a vector
bundle) (here, G1(Rn) is the set of all 1subspaces of Rn, i.e. the
projective space of Rn) ???
P.S. For n=2, it is not orientable, because it is the moebius strip.

I'll use the notation RP^(n1) for G1(Rn), and refer to it as "real
projective (n1) space".
Your space, which I'll denote X, is orientable iff n is odd.
To prove this, note that the sphere bundle of the canonical line
bundle over RP^(n1) has total space equal to the (n1) dimensional
sphere, S^(n1), and its projection is the canonical double cover
f: S^(n1) > RP^(n1). Further, if this projection is used to
attach an ncell D^n to RP^(n1), yielding the complex:
D^n \cup_f RP^(n1)
then the result is RP^n, alias real projective nspace. The normal
bundle of RP^(n1) in RP^n is the canonical line bundle over RP^(n1),
so the space X is homeomorphic to a tubular neighborhood of RP^(n1)
in RP^n. I'll identify X with that tubular neighborhood.
If RP^n is orientable, then X (being an open submanifold) must also be,
and if X is orientable, then RP^n must be orientable, since RP^n \ X is
contractible (it's a disc), .
Finally, RP^n is easily shown to be orientable iff n is odd: RP^n is
the quotient of S^n by the antipodal map, and that map is orientation
preserving (allowing one to produce an orientation on the quotient) iff
the map x > x is orientationpreserving on R^(n+1).
Dale 

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Ruitao Zhang science forum beginner
Joined: 24 Mar 2005
Posts: 1

Posted: Thu Mar 24, 2005 9:51 pm Post subject:
Re: de Finetti's theorem



de Finetti theorem said that if an infinite sequence is exchangeable
then there exist a unique probability measure such that the de finetti
hold. Is this probability measure the prior measure. In Bayesian we
choose start from any prior measure. Here it seems that the prior is
unique. I don't understand it.
For infinite population with exchangeability, we can use relative
frequency of an event to estimate the the probability of this event.
But, For the finite population, why we need exchangeability?
Thanks very much for your help
Ruitao 

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Ettore Aldrovandi science forum beginner
Joined: 24 Mar 2005
Posts: 1

Posted: Thu Mar 24, 2005 9:51 pm Post subject:
a post that didn't appear



In article <ctbmgs$b8s$1@news.ks.uiuc.edu>,
Urs Schreiber <Urs.Schreiber@uniessen.de> wrote:
Quote:  "John Baez" <baez@math.removethis.ucr.andthis.edu> schrieb im Newsbeitrag
news:ctas57$1p2$1@news.ks.uiuc.edu...
2bundles are meant to be an alternative to gerbes: although
I've done my best to hide it above, a gerbe is really more
like a categorified *sheaf* than a bundle. And, just as a
bundle has a sheaf of sections, we're hoping that a 2bundle
has a stack of sections, which in certain cases will be a
gerbe. That's one of the things we need to figure out,
though.
First, by definition, a gerbe is a stack is a fibered category. The concept
"fibered category" is a categorification of the concept "presheaf". But not
its full categorification. I am wondering if the full categorification of
the concept "presheaf" has been studied before, and under which name.
More precisely, what I am talking about is this:
A presheaf over a topological space X is a morphism in Cat,
namely a (contravariant) functor from the category O(X) of open
subsets of X to Set.

Hi, you are maybe rigidifying the starting situation a bit too
much? A gerbe is a locally nonempty and locally connected stack
in groupoids. In particular, it is a fibered category. Now, this
does not make it a contravariant functor in Cat, because the
restriction functors commute only up to natural
transformation. If memory serves, this is called a
"pseudofunctor" in SGA1. "Laxfunctor" is also used, I think.
Note also that the "base" can be any site, I believe. Of course
so is your O(X) when you consider it as a category. At any rate,
what I'm saying is this:
if p: G > X is a fibered category over X, say G is a gerbe, but
id doesn't matter here, if a, b, c are objects of X, then you
have the corresponding fiber categories G(a), G(b), G(c).If
i:b>a
is a morphism in X, then there is a corresponding "restriction"
functor
i^*: G(a) > G(b).
and if now j: c > b is another morphism with the corresponding
functor
j^*: G(b) > G(c)
then there is only a natural transformation
(ij)^* ==> j^*i^*
between the two resulting functors from G(a) to G(c).

Ettore Aldrovandi
Department of Mathematics http://www.math.fsu.edu/~ealdrov
Florida State University aldrovandi at math.fsu.edu
Tallahassee, FL 323064510, USA +1 (850) 6449717 (FAX: 4053) 

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Peter Spellucci science forum Guru
Joined: 29 Apr 2005
Posts: 702

Posted: Thu Mar 24, 2005 9:51 pm Post subject:
Re: Defining an Infinitely Differentiable Function



In article <1107281374.556304@athnrd02>,
Ioannis <jgal@ath.forthnet.gr> writes:
Quote:  I am in possession of a sequence of functions {S(m,z):NxD>C}_{m \in N},
and their series valid in some domain D < C:
S(1,z)=sum(a_{1,i}*z^i,i=0..+inf),
S(2,z)=sum(a_{2,i}*z^i,i=0..+inf),
...
S(m,z)=sum(a_{m,i}*z^i,i=0..+inf),
...
I am interested in defining (in D), S(r,z), r \in R (resp S(w,z) w \in
C by analytic continuation), so that S(r,z) (resp S(w,z)) is C^{oo} with
respect to r (resp w).
The a_{m,n} obey a recursion:
a_{m,n}={1, if n=0,
1/n!, if m=1,
{sum(j*a_{m,nj}*a_{m1,j1},j=1..n)}/n, otherwise}
This recursion doesn't seem to have a closed form. If it had, I could
use the analytic continuation (for fixed n_0) of its closed form, to
define S(w,z).
I tried using various functions (including polynomial and linear
interpolation) to interpolate between the coefficients {a_{m,n_0}_{m \in
N} for fixed n_0, but the resultant function's derivative is always
discontinuous at m \in N.
So that I don't waste any more time with this, my question is, is there
any guarantee that S(w,z) can be defined to be C^{oo} with respect to r
(or w), by interpolating between the coefficients (vertically) or should
I be looking at other methods?
Thanks much in advance,

I. N. G.  http://users.forthnet.gr/ath/jgal/

only a vague idea: consider the mapping [1,infty]>[0,1]
z=1/(x+1)
for the variable "r" (i.e. the "m" in a_{m,n})
apply this to the "function" a_{m,n} with respect to the variable m, n
fixed.
define a sequence of finite fourierseries interpolating the first 1,...M
a{i,n}
for every n separately and then analyze the convergence of the function
sequence defined by this. then consider the convergence of the corresponding
S(f(m,n),z) series defined by this, f(r,n) being the fourierseries with
the variable "m", n fixed, (after backtransformation to [1,infty[)
and f(r,n) replacing the coefficient a_{m,n}
your recursion looks as if the a_{m,n} decay fast enough in order to be able to
show that the final series obtained this way is absolutely convergent.
hth
peter 

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Robin Chapman science forum Guru Wannabe
Joined: 25 Mar 2005
Posts: 254

Posted: Thu Mar 24, 2005 9:51 pm Post subject:
Re: Orthogonal latin squares of even side



Simone Severini wrote:
Quote:  Would you explain to me a moreorless general (possibly simple!)
method to construct orthogonal latin squares of even side?

At
http://wwwmath.cudenver.edu/~wcherowi/courses/m6406/cslne.html
there is a short account by Bill Cherowitzo of a construction by Zhu Lie.

Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
"Elegance is an algorithm"
Iain M. Banks, _The Algebraist_ 

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Axel Vogt science forum addict
Joined: 03 May 2005
Posts: 93

Posted: Thu Mar 24, 2005 9:51 pm Post subject:
Re: The curse of dimensionality for integration



On Thu, 19 Feb 2004 06:38:52 +0000 (UTC), Greg Kuperberg wrote:
Quote:  In article <4033DA60.7060503@univie.ac.at>,
Arnold Neumaier <Arnold.Neumaier@univie.ac.at> wrote:
Numerically, integration is simpler than differentiation in one
dimension, but in higher dimension, integration suffers from the
curse
of dimensionality while differntiation doesn't. In particular, it is
very hard to get accurate integrals in dimensions >100, say.
Readers may be interested in my new paper in which I fight the
curse of dimensionality for numerical integration in high dimensions:
a
href="http://front.math.ucdavis.edu/math.NA/0402047">http://front.math.ucdavis.edu/math.NA/0402047</a
Granted, it's impossible to completely defeat the curse. I derive
good methods for functions that are wellapproximated by lowdegree
polynomials.

Is there an example available to look at for computing a cumulative
multivariate normal distribtuion?

use mail ät axelvogt dot de 

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John Ryskamp science forum beginner
Joined: 24 Mar 2005
Posts: 1

Posted: Thu Mar 24, 2005 9:51 pm Post subject:
Re: Brouwer's characterization of Cantor set



The most important response of Brouwer to Cantor was Brouwer's
formulation of an infinite ordinal number. He described this in his
1912 lecture. However, it is based on that idea that Cantor proved
the wellordering of the ordinal numbers. As Garciadiego has shown,
not only did Cantor not do so, but also, he never claimed to have done
so, and never used the term infinite ordinal number. The term
infinite ordinal number has no meaning and you should examine your
interests in light of what Garciadiego has to say about the math
history of Brouwer's era.
On Thu, 09 Dec 2004 18:09:40 +0000,
=?ISO88591?Q?Jos=E9_Carlos_Santos?= wrote:
Quote:  On 08122004 12:46, Jorge Buescu wrote:
Brouwer has given an equivalent characterization of the Cantor set
as
a perfect, totally disconnected, compact Hausdorff space with a
countable
base of clopen subsets.
However, I can't seem to find a suitable reference for this fact
(MathSciNet does not extend that far). But this is probably on
some
appropriate General Topology books. Can anyone point one out?
There's a proof in Willi Rinow's Lehrbuch der Topologie. It's the
last
theorem of section 24.
Best regards,
Jose Carlos Santos 


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G A Edgar science forum beginner
Joined: 24 Mar 2005
Posts: 1

Posted: Thu Mar 24, 2005 9:51 pm Post subject:
Re: Brouwer's characterization of Cantor set



In article <cvkodf$vq5$1@news.ks.uiuc.edu>, John Ryskamp
<philneo2001@yahoo.com> wrote:
Quote:  The most important response of Brouwer to Cantor was Brouwer's
formulation of an infinite ordinal number. He described this in his
1912 lecture. However, it is based on that idea that Cantor proved
the wellordering of the ordinal numbers. As Garciadiego has shown,
not only did Cantor not do so, but also, he never claimed to have
done
so, and never used the term infinite ordinal number.

???
"Ueber unendliche, lineare Punktmannigfaltigkeiten, 5"
Math. Annalen 21 (1883) 545586
available online:
http://wwwgdz.sub.unigoettingen.de/cgibin/digbib.cgi?PPN235181684_0021
I guess you could say Cantor did not use the term "infinite ordinal
number"
because, writing in German, he used "unendlichen Zahlen" (see page
547)...
But somehow I do not think that is what Garciadiego had in mind. 

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