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How real are the "Virtual" partticles?
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George Baloglou
science forum beginner


Joined: 09 May 2005
Posts: 7

PostPosted: Thu Sep 22, 2005 9:56 am    Post subject: Re: This Week's Finds in Mathematical Physics (Week 221) Reply with quote

In article <dgrghh$63i$1@dizzy.math.ohio-state.edu>,
John Baez <baez@math.removethis.ucr.andthis.edu> wrote:
Quote:

Some corrections and addenda:

[snip]

Quote:
Archimedes did more work on calculus than previously believed!
We know this now because a manuscript of his that had been erased
and written over has recently been read with the help of a
synchrotron X-ray beam!

http://www.mlahanas.de/Greeks/ArchimedesPal.htm
http://news-service.stanford.edu/news/2005/may25/archimedes-052505.html

This manuscript also reveals for the first time that he did work on
combinatorics:

http://www.mlahanas.de/Greeks/ArchimedesComb.htm

For a modern/combinatorial understanding of Archimedes' work on the puzzle
STOMACH(ION), one should visit http://www.math.ucsd.edu/~fan/stomach/ ;
that was also front page news on the Sunday New York Times of 12/14/03!

George Baloglou
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anz3
science forum beginner


Joined: 21 Sep 2005
Posts: 12

PostPosted: Thu Sep 22, 2005 8:00 pm    Post subject: Re: extension of Minkowski's inequality Reply with quote

Thank you very much for your answer. Since I am still searching for a
prove of the validity of Minkowski's inequality with parameter \infty
< p < 0, I would appreciate very much if someone could tell me where
in the literature that can be found. Again, it would be great if the
result were formulated in the terminology of random variables.
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martin cohen
science forum Guru Wannabe


Joined: 18 May 2005
Posts: 104

PostPosted: Fri Sep 23, 2005 1:00 pm    Post subject: Re: extension of Minkowski's inequality Reply with quote

anz3 wrote:

Quote:
Thank you very much for your answer. Since I am still searching for a
prove of the validity of Minkowski's inequality with parameter \infty
p < 0, I would appreciate very much if someone could tell me where
in the literature that can be found. Again, it would be great if the
result were formulated in the terminology of random variables.

Both Hardy, Littlewood, and Polya's "Inequalities"

and Beckenbach and Bellman's "Inequalities" have the theorems
and the proofs. Just look in the table of contents.
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Ali Taghavi
science forum addict


Joined: 14 May 2005
Posts: 73

PostPosted: Sat Sep 24, 2005 1:30 pm    Post subject: Re: Which TVS is compatible with Lie bracket? Reply with quote

Quote:
In fact the Example of Robert Israel shows that the following
question 1 has negative answer,since the spectrum is unbounded,But

what about the following second question 2(search for TVS structur
compatible to the data?):

in the following V is a linear space and T is a linear map from V to V

Question 1:Is there a norm on V such that T is a bounded operator

question 2:is there a topological vector space structure on V such
that T is continious linear map?

Ali Taghavi
Quote:

According to my example, there's no possibility of
such a norm on
any subspace that includes 1 and exp(kx) for
arbitrarily large k.



Robert Israel
israel@math.ubc.ca
Department of Mathematics
http://www.math.ubc.ca/~israel
University of British Columbia Vancouver,
BC, Canada
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Anton Deitmar
science forum beginner


Joined: 25 Jul 2005
Posts: 24

PostPosted: Mon Sep 26, 2005 6:30 pm    Post subject: Re: Which TVS is compatible with Lie bracket? Reply with quote

Quote:
in the following V is a linear space and T is a
linear map from V to V

Question 1:Is there a norm on V such that T is a
bounded operator

Counterexample: let V have the basis e_1, e_2, e_3, ...
Define T by

T(e_j) = j e_j

Then for any norm N:

N(T(e_j)) = j N(e_j)

so T is not bounded.

Quote:

question 2:is there a topological vector space
structure on V such
that T is continious linear map?

Yes, let N be a norm on V and set

s_k(v) = N(T^k(v))

Then the family s_k, k=0,1,2,...
of seminorms defines a topology on V and one has

s_k(T(v)) = s_{k+1}(v),

hence T is continuous wrt this topology.

Cheers,
Anton
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M.A.Fajjal
science forum Guru Wannabe


Joined: 13 Sep 2005
Posts: 115

PostPosted: Tue Oct 11, 2005 1:20 am    Post subject: Re: Solving solvable QUINTICS using two sextics Reply with quote

Quote:
Hello all,

For those interested in the quintic, this paper might
be of some help:

"An Easy Way To Solve Solvable Quintics Using Two
Sextics"

ABSTRACT: Using a method initially developed by
George Young (1819-
1889), Arthur Cayley (1821-1895), and later by George
Watson (1886-
1965), an explicit quartic is constructed to enable
the solution in
radicals of a quintic when it is a solvable equation.
Not one, but
two sextic resolvents are derived which are important
to forming the
coefficients of this quartic. Certain difficulties
and their
solutions as well as a novel consequence to the
method are also
addressed in this paper.

Mathematics Subject Classification. Primary: 12E12;
Secondary: 12F10.

http://www.geocities.com/titus_piezas/

Just click on the link in the website to the pdf
file.


Sincerely,

Titus (tpiezasIII@uap.edu.ph -> remove III for email)

Thank you for your simple method for solving solvable quintics by using two sextics. However, I have faced some problems when using your method for solving quintics. The following example shows these problems.

x^5-10102/5*x^3+1368821/25*x^2-67067178/125*x+5658741371/3125=0

c=-10102/50
d=1368821/250
e=-67067178/625
f=5658741371/3125

p=0
The resolvent quartic

z^4+5658741371/3125*z^3+14030242707268383381/9765625*z^2+18604015291701468087887634121/30517578125*z +10808705528596191952313957779475163001/95367431640625=0

ans =

[ -5658741371/12500+5051/100*5523005^(1/2)+5051/2500*(-28816410650-11203210*5523005^(1/2))^(1/2)]
[ -5658741371/12500+5051/100*5523005^(1/2)-5051/2500*(-28816410650-11203210*5523005^(1/2))^(1/2)]
[ -5658741371/12500-5051/100*5523005^(1/2)+5051/2500*(-28816410650+11203210*5523005^(1/2))^(1/2)]
[ -5658741371/12500-5051/100*5523005^(1/2)-5051/2500*(-28816410650+11203210*5523005^(1/2))^(1/2)]

Which root shall be assigned for z1, z2, z3, and z4? Why?

As all z1, z2, z3 and z4 are complex numbers

each of z1^(1/5), z2^(1/5), z3^(1/5) and z4^(1/5) shall have 5 possible values. Which value for each z_i shall be taken? Why?

I think, it should be some assigned formulae to determine each z_i and each z_i^(1/5)

Best regards
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Titus Piezas III
science forum Guru Wannabe


Joined: 10 Mar 2005
Posts: 102

PostPosted: Fri Oct 14, 2005 11:18 am    Post subject: Re: Solving solvable QUINTICS using two sextics Reply with quote

M.A.Fajjal wrote:
Quote:
Hello all,

For those interested in the quintic, this paper might
be of some help:

"An Easy Way To Solve Solvable Quintics Using Two
Sextics"

ABSTRACT: Using a method initially developed by
George Young (1819-
1889), Arthur Cayley (1821-1895), and later by George
Watson (1886-
1965), an explicit quartic is constructed to enable
the solution in
radicals of a quintic when it is a solvable equation.
Not one, but
two sextic resolvents are derived which are important
to forming the
coefficients of this quartic. Certain difficulties
and their
solutions as well as a novel consequence to the
method are also
addressed in this paper.

Mathematics Subject Classification. Primary: 12E12;
Secondary: 12F10.

http://www.geocities.com/titus_piezas/

Just click on the link in the website to the pdf
file.


Sincerely,

Titus (tpiezasIII@uap.edu.ph -> remove III for email)

Thank you for your simple method for solving solvable quintics by using two sextics. However, I have faced some problems when using your method for solving quintics. The following example shows these problems.

x^5-10102/5*x^3+1368821/25*x^2-67067178/125*x+5658741371/3125=0

c=-10102/50
d=1368821/250
e=-67067178/625
f=5658741371/3125

p=0
The resolvent quartic

z^4+5658741371/3125*z^3+14030242707268383381/9765625*z^2+18604015291701468087887634121/30517578125*z +10808705528596191952313957779475163001/95367431640625=0

ans =

[ -5658741371/12500+5051/100*5523005^(1/2)+5051/2500*(-28816410650-11203210*5523005^(1/2))^(1/2)]
[ -5658741371/12500+5051/100*5523005^(1/2)-5051/2500*(-28816410650-11203210*5523005^(1/2))^(1/2)]
[ -5658741371/12500-5051/100*5523005^(1/2)+5051/2500*(-28816410650+11203210*5523005^(1/2))^(1/2)]
[ -5658741371/12500-5051/100*5523005^(1/2)-5051/2500*(-28816410650+11203210*5523005^(1/2))^(1/2)]

Which root shall be assigned for z1, z2, z3, and z4? Why?

As all z1, z2, z3 and z4 are complex numbers

each of z1^(1/5), z2^(1/5), z3^(1/5) and z4^(1/5) shall have 5 possible values. Which value for each z_i shall be taken? Why?

I think, it should be some assigned formulae to determine each z_i and each z_i^(1/5)

Best regards


Fajjal,

Wow, I didn't think anyone would reply to this post after a year.
Anyway, as Dummit pointed out in his paper, this is the quintic "casus
irreducibilis" analogous to the cubic version, when all roots are real.

The formula (just like Cardano's) remains true generically though
numerically, when the quintic has all roots real one has to properly
determine which 5th roots of the 4 complex roots of its resolvent go
together, as you pointed out.

Dummit had a similar complication with his method though he got around
that by using an "ordering condition" and letting the 5th roots satisfy
a certain equation, similar to what you did in another post.

To make the method as simple as possible I ignored this aspect though,
with some tweaking, it can address this as well.

P.S. You really should write some formal paper on your work on quintics
in some easy-to-read font and post it on a website. Others will
appreciate it. (For some reason, there's always someone wanting to know
about quintics.) ASCII is so discouraging to read.

--Titus
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Ali Taghavi
science forum addict


Joined: 14 May 2005
Posts: 73

PostPosted: Thu Oct 20, 2005 1:45 am    Post subject: Re: IS quantum mechanics a limit cycle theory?? Reply with quote

hello
let we have a classical hamiltonian H,and we quantize it.
I want to learn some relations between the behavior of solutions of X_H(the classic hamiltonian vector field) and some operetor theoretic invariants of the quantum operatores.In particular a nice interprewtation of the number of closed orbits of X_H in quantum language!
Thank you for your suggestions for some deep references
Ali Taghavi

Quote:
Hi
I Am Intersted in the Hilbert 16th problem which main
object is "Limit
Cycle"!
Last Year I Found in the web a paper by Cetto and De
La Penna :with
The Title "Is Quantum Mechanics A Limit Cycle
Theory?"
this paper is available in Mathscinet.
I Invite you to review this paper and discuss on a
possible relation
between Hilbert 16th Problem And Mathematical Aspect
Of QM,
for begining:Let We Have A Planner Vector field
L(Lienard Equation )
as follow:
x'=y-F(x) y'=-x where F is A non even_polynomial ,we
are intersted in
the number of Limit Cycles Of L,consider the
following two
questions:(Assume F'(0) is not zero)
1)Does there exist a correspondence between {closed
orbites} of L and
{closed orbits} of 4 dimensional (classical)
Hamiltonian (y-F(x))z-xw?
please see also the similar question in :

http://www.arxiv.org/abs/math.CA/0409594

2)Let's Quantize the above 4 dimensional
hamiltonian:x,y stand for
operators multiplication by x and y,while z,w stand
for partial
derivative with respect to x ,y resp.
What Is The Quantum interpretations for (the number
of) closed orbits
of classical Hamiltonian
(y-F(x))z-xw?
Thank you
Ali Taghavi
Iran
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Anton Deitmar
science forum beginner


Joined: 25 Jul 2005
Posts: 24

PostPosted: Thu Oct 20, 2005 1:35 pm    Post subject: Re: Presheaf as functor Reply with quote

Being a presheaf, it is a functor from the category of opens in X to the category of abelian groups.
The O(U)-module structure on F(U) is an add-on structure.
This is why people rather speak of O_X-modules.
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Ali Taghavi
science forum addict


Joined: 14 May 2005
Posts: 73

PostPosted: Mon Oct 24, 2005 12:30 pm    Post subject: Re: IS quantum mechanics a limit cycle theory?? Reply with quote

Due to material of my question in "Is quantum mechanics a limit cycle
theory" , it seems that I should ask, in priori, for an example of a
classical hamiltonian H with a finite (but non zero) number of closed
orbits (of course it is impossible in 2 dim.) If it is impossible to
present an example of hamiltonian with a finite number of closed
orbits, so the quantum interpretation for this "number" would be
meaningless!


Quote:
hello
let we have a classical hamiltonian H,and we quantize
it.
I want to learn some relations between the behavior
of solutions of X_H(the classic hamiltonian vector
field) and some operetor theoretic invariants of the
quantum operatores.In particular a nice
interprewtation of the number of closed orbits of X_H
in quantum language!
Thank you for your suggestions for some deep
references
Ali Taghavi

Hi
I Am Intersted in the Hilbert 16th problem which
main
object is "Limit
Cycle"!
Last Year I Found in the web a paper by Cetto and
De
La Penna :with
The Title "Is Quantum Mechanics A Limit Cycle
Theory?"
this paper is available in Mathscinet.
I Invite you to review this paper and discuss on a
possible relation
between Hilbert 16th Problem And Mathematical
Aspect
Of QM,
for begining:Let We Have A Planner Vector field
L(Lienard Equation )
as follow:
x'=y-F(x) y'=-x where F is A non even_polynomial
,we
are intersted in
the number of Limit Cycles Of L,consider the
following two
questions:(Assume F'(0) is not zero)
1)Does there exist a correspondence between {closed
orbites} of L and
{closed orbits} of 4 dimensional (classical)
Hamiltonian (y-F(x))z-xw?
please see also the similar question in :

http://www.arxiv.org/abs/math.CA/0409594

2)Let's Quantize the above 4 dimensional
hamiltonian:x,y stand for
operators multiplication by x and y,while z,w stand
for partial
derivative with respect to x ,y resp.
What Is The Quantum interpretations for (the number
of) closed orbits
of classical Hamiltonian
(y-F(x))z-xw?
Thank you
Ali Taghavi
Iran

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Han de Bruijn
science forum Guru


Joined: 18 May 2005
Posts: 1285

PostPosted: Mon Oct 24, 2005 2:30 pm    Post subject: Re: minimum volume ellipse Reply with quote

John wrote:

Quote:
I am looking for a solution that works in high dimensions.
Most of these solutions are exponential in the dimension.

How about trying to generalize the theory at:

http://huizen.dto.tudelft.nl/deBruijn/grondig/crossing.htm#BE

to multiple dimensions? Involving a quadratic form with the mean values
of the points as a midpoint, and the inverse of a matrix with variances
giving the coefficients.

Must be not too bad. Just my 5 cents worth ...

Han de Bruijn

..
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joe.shmoe.joe.shmoe
science forum beginner


Joined: 19 Oct 2005
Posts: 7

PostPosted: Mon Oct 24, 2005 5:00 pm    Post subject: Re: Presheaf as functor Reply with quote

Thanks for the replies. This sounds like a good way to do it, but it's
not exactly what I had in mind. I started thinking about this after
reading the definition of a presheaf in many textbooks, along the lines
of:
"On a topological space X, a presheaf F of sets (or abelian groups,
rings, etc) is an assignment for each open subset..." etc.
Clearly, it seems beneficial to prove statements about presheaves in
the general context of "contravariant functors from opens(X) to C" for
some C. Then looking at stalks, for example, works the same for all
categories in which direct limits exist. Or another example -
constructing the sheaf associated to a presheaf, which in the case of
sets, abelian groups etc., gives you a (pre)sheaf with values in the
same category you started with. This is the sort of thing I'm trying to
achieve with some degree of generality.

One could add slightly modify the definition of a presheaf, starting
with a fixed "base" functor R: Opens(X) --> B, and a "projection"
functor P:C --> B, and then saying a presheaf into C ("relative to R
and P") is a functor F:Opens(X)--> C such that PF=R. However, this
does make things a lot more cumbersome, and I have not seen this sort
of thing done anywhere.

Any thoughts on this?
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Robert B. Israel
science forum Guru


Joined: 24 Mar 2005
Posts: 2151

PostPosted: Mon Oct 24, 2005 5:30 pm    Post subject: Re: IS quantum mechanics a limit cycle theory?? Reply with quote

Ali Taghavi wrote:
Quote:
Due to material of my question in "Is quantum mechanics a limit cycle
theory" , it seems that I should ask, in priori, for an example of a
classical hamiltonian H with a finite (but non zero) number of closed
orbits (of course it is impossible in 2 dim.) If it is impossible to
present an example of hamiltonian with a finite number of closed
orbits, so the quantum interpretation for this "number" would be
meaningless!

Consider a particle moving in 3 dimensions in the potential
V(x,y,z) = (1 - x^2 - y^2)^2 z + z^3 + x^2 + y^2

Note that dV/dz > 0 except on the circle x^2 + y^2 = 1, z = 0, so
the only possible closed orbits are on that circle. On that circle
there are closed orbits in both directions.

Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
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Han de Bruijn
science forum Guru


Joined: 18 May 2005
Posts: 1285

PostPosted: Tue Oct 25, 2005 2:30 pm    Post subject: Re: IS quantum mechanics a limit cycle theory?? Reply with quote

israel@math.ubc.ca wrote:

Quote:
Consider a particle moving in 3 dimensions in the potential
V(x,y,z) = (1 - x^2 - y^2)^2 z + z^3 + x^2 + y^2

Note that dV/dz > 0 except on the circle x^2 + y^2 = 1, z = 0, so
the only possible closed orbits are on that circle. On that circle
there are closed orbits in both directions.

IMHO this cannot have any physical relevance, because these orbits are
highly _unstable_. Am I wrong?

Han de Bruijn

..
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Ali Taghavi
science forum addict


Joined: 14 May 2005
Posts: 73

PostPosted: Tue Oct 25, 2005 2:30 pm    Post subject: Re: IS quantum mechanics a limit cycle theory?? Reply with quote

the domain of hamiltonian is even dimensional space! But can Morse
theory help to my question, namely is dynamic fix if we do not pass a
critical value (However I prefer to not change the main subject of my
question, a possible relation between limit cycle theory and
quantization)

Quote:
Consider a particle moving in 3 dimensions in the
potential
V(x,y,z) = (1 - x^2 - y^2)^2 z + z^3 + x^2 + y^2

Note that dV/dz > 0 except on the circle x^2 + y^2 =
1, z = 0, so
the only possible closed orbits are on that circle.
On that circle
there are closed orbits in both directions.

Robert Israel
israel@math.ubc.ca
Department of Mathematics
http://www.math.ubc.ca/~israel
University of British Columbia Vancouver,
BC, Canada
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