Author 
Message 
alain verghote science forum Guru Wannabe
Joined: 29 Apr 2005
Posts: 293

Posted: Thu Jul 28, 2005 3:00 pm Post subject:
Re: " How do you build an involutive operator? "



Dear Robert,
Many thanks for your reply.
Two points:
1° ) your periodic operator is very closed to
homographic function h(x)=(x+tan(t) ) / (1x*tan(t))
with tan(n*t) = 0 and ad hoc cycle n .
2°) Involutive operators:
I do agree with your explanation and remark f(0)=0
As we may construct involutive function just using
(x,f) symmetry ,example:
3/(f2) + 3/(x2) =40 > f(x) ...
I constructed the given example upon:
(a +Dphi)*(a +Df) = c , D for d/dx ,
Dphi for D phi(f)
Friendly,
Alain. 

Back to top 


Guest

Posted: Thu Jul 28, 2005 3:00 pm Post subject:
Re: Separable Linear Order



William Elliot wrote:
Quote:  Given a linear order L with the usual interval topology,
how does one show when L is separable, L is Lindelof?

Let C be any open cover of L; we may assume that the members of C are
open intervals (a, b). Define an equivalence relation on L so that,
when x < y, the points x and y are equivalent just in case the closed
interval [x, y] is covered by countably many members of C. Each
equivalence class is open, since it is the union of some subcollection
of C. So the equivalence classes form a collection of disjoint nonempty
open subsets of L. Since L is separable, it follows that there are only
countably many equivalence classes. To finish the proof, we have to
show that each equivalence class is covered by countably many members
of C.
Quote:  From the separability of L, it follows that L cannot have a subset of
order type omega_1. Hence each equivalence class must have a countable 
cofinal subset, and (dually) a countable coinitial subset. Hence each
equivalence class is the union of countably many closed intervals. By
the definition of the equivalence relation, each closed interval
(contained in an equivalence class) is covered by countably many
members of C. That does it, I think. 

Back to top 


Aldar CF. Chan science forum addict
Joined: 30 Apr 2005
Posts: 52

Posted: Fri Jul 29, 2005 2:00 pm Post subject:
Re: Counting partition patterns of a set



"Gerry Myerson" <gerry@maths.mq.edi.ai.i2u4email> wrote in message
news:dcartn$bhl$1@news.ks.uiuc.edu...
Quote:  In article <dc85pf$g12$1@news.ks.uiuc.edu>,
"Aldar CF. Chan" <aldar@comm.utoronto.ca> wrote:
Suppose I have a set X and partition it in different ways, say
P_1, P_2, ..., P_i,..., P_m. That is, each P_i is a distinct
partition of X. For any randomly selected subset of X, say
A \subset X, if I want to make sure X\A is equal to some
union of Y_i's where each Y_i is an element of the union of
P_1, ..., P_m. What is the minimum possible m for this to be
fulfilled? And how such a collection of partition patterns can
be found?
Perhaps I'm confused, but it appears to me
that you need just one partitiion,
the partition into singletons.
Any subset of X is a union of singletons.

Gerry Myerson (gerry@maths.mq.edi.ai) (i > u for email)

I forgot to add another side question. Suppose we want to keep m to
be smaller than some maximum say M, what's the average number of
Y_i needed then? 

Back to top 


Aldar CF. Chan science forum addict
Joined: 30 Apr 2005
Posts: 52

Posted: Fri Jul 29, 2005 2:00 pm Post subject:
Re: Counting partition patterns of a set



"Gerry Myerson" <gerry@maths.mq.edi.ai.i2u4email> wrote in message
news:dcartn$bhl$1@news.ks.uiuc.edu...
Quote:  In article <dc85pf$g12$1@news.ks.uiuc.edu>,
"Aldar CF. Chan" <aldar@comm.utoronto.ca> wrote:
Suppose I have a set X and partition it in different ways, say
P_1, P_2, ..., P_i,..., P_m. That is, each P_i is a distinct
partition of X. For any randomly selected subset of X, say
A \subset X, if I want to make sure X\A is equal to some
union of Y_i's where each Y_i is an element of the union of
P_1, ..., P_m. What is the minimum possible m for this to be
fulfilled? And how such a collection of partition patterns can
be found?
Perhaps I'm confused, but it appears to me
that you need just one partitiion,
the partition into singletons.
Any subset of X is a union of singletons.

Gerry Myerson (gerry@maths.mq.edi.ai) (i > u for email)

Yes, you're right. I forgot to say I want the average number of Y_i
needed over all possible A to be minimum as well. Is that easy this
time? 

Back to top 


Aldar CF. Chan science forum addict
Joined: 30 Apr 2005
Posts: 52

Posted: Fri Jul 29, 2005 2:00 pm Post subject:
Re: Counting partition patterns of a set



"Kloeckner Benoît" <hotel.des@invalides.fr> wrote in message
news:dcarto$bhq$1@news.ks.uiuc.edu...
Quote:  Aldar CF. Chan wrote:
Suppose I have a set X and partition it in different ways, say
P_1, P_2, ..., P_i,..., P_m. That is, each P_i is a distinct
partition of X. For any randomly selected subset of X, say
A \subset X, if I want to make sure X\A is equal to some
union of Y_i's where each Y_i is an element of the union of
P_1, ..., P_m. What is the minimum possible m for this to be
fulfilled? And how such a collection of partition patterns can
be found?
If you do not impose any restriction on your partition, m=1 is
minimum: let P_1={{x}; x in X} and it's over.
So I suppose you impose a restriction.
If you want each of your partitions to have two elements, then
m=Card(X) is minimal and you can use the family (P_x = {{x},{X\x}}) for
the following reason.
For any x in X, let A_x be X\x. Then you want {x}=X\Ax to be a term
of one of your partition, thus you must include P_x in your family of
partition. Conversly, with all P_x's you can rebuild any subset of X.

After thinking more carefully, I guess having both m and the average number
of
Y_i minimum do not make much sense. What I have in mind is to find the
minimum
m such that the cover (that is, the number of Y_i needed for any particular
A) would
be minimum in some sense but I am not sure how this can be formulated. 

Back to top 


K. P. Hart science forum beginner
Joined: 03 May 2005
Posts: 12

Posted: Fri Jul 29, 2005 2:00 pm Post subject:
Re: Separable Linear Order



fred.galvin@gmail.com wrote:
Quote:  William Elliot wrote:
Given a linear order L with the usual interval topology,
how does one show when L is separable, L is Lindelof?
Let C be any open cover of L; we may assume that the members of C are
open intervals (a, b). Define an equivalence relation on L so that,
when x < y, the points x and y are equivalent just in case the closed
interval [x, y] is covered by countably many members of C. Each
equivalence class is open, since it is the union of some subcollection
of C. So the equivalence classes form a collection of disjoint nonempty
open subsets of L. Since L is separable, it follows that there are only
countably many equivalence classes. To finish the proof, we have to
show that each equivalence class is covered by countably many members
of C.
From the separability of L, it follows that L cannot have a subset of
order type omega_1. Hence each equivalence class must have a countable
cofinal subset, and (dually) a countable coinitial subset.
More directly: 
let D be a countable dense set in L that includes the end points, if any.
Then L = bigcup{[a,b]: a,b in D; a<b} writes L as a union of countably
many closed intervals, each of which is covered by countably many O's.
KP

EMAIL: K.P.Hart@EWI.TUDelft.NL PAPER: Faculteit EWI
PHONE: +31152784572 TU Delft
FAX: +31152787245 Postbus 5031
URL: http://fa.its.tudelft.nl/~hart 2600 GA Delft
the Netherlands
.. 

Back to top 


Fred Galvin science forum beginner
Joined: 31 Jul 2005
Posts: 21

Posted: Sun Jul 31, 2005 3:30 pm Post subject:
Re: Separable Linear Order



fred.galvin@gmail.com wrote:
Quote:  William Elliot wrote:
Given a linear order L with the usual interval topology,
how does one show when L is separable, L is Lindelof?
Let C be any open cover of L; we may assume that the members of C are
open intervals (a, b). Define an equivalence relation on L so that,
when x < y, the points x and y are equivalent just in case the closed
interval [x, y] is covered by countably many members of C. Each
equivalence class is open, since it is the union of some subcollection
of C. So the equivalence classes form a collection of disjoint nonempty
open subsets of L. Since L is separable, it follows that there are only
countably many equivalence classes. To finish the proof, we have to
show that each equivalence class is covered by countably many members
of C.
From the separability of L, it follows that L cannot have a subset of
order type omega_1. Hence each equivalence class must have a countable
cofinal subset, and (dually) a countable coinitial subset. Hence each
equivalence class is the union of countably many closed intervals. By
the definition of the equivalence relation, each closed interval
(contained in an equivalence class) is covered by countably many
members of C. That does it, I think.

PS. Note that, instead of assuming that L is separable, you can get by
with the weaker assumption that L satisfies the "countable chain
condition", i.e., every collection of disjoint open sets is countable. 

Back to top 


Fred Galvin science forum beginner
Joined: 31 Jul 2005
Posts: 21

Posted: Sun Jul 31, 2005 3:30 pm Post subject:
Re: Separable Linear Order



K. P. Hart wrote:
Quote:  fred.galvin@gmail.com wrote:
William Elliot wrote:
Given a linear order L with the usual interval topology,
how does one show when L is separable, L is Lindelof?
Let C be any open cover of L; we may assume that the members of C are
open intervals (a, b). Define an equivalence relation on L so that,
when x < y, the points x and y are equivalent just in case the closed
interval [x, y] is covered by countably many members of C. Each
equivalence class is open, since it is the union of some subcollection
of C. So the equivalence classes form a collection of disjoint nonempty
open subsets of L. Since L is separable, it follows that there are only
countably many equivalence classes. To finish the proof, we have to
show that each equivalence class is covered by countably many members
of C.
From the separability of L, it follows that L cannot have a subset of
order type omega_1. Hence each equivalence class must have a countable
cofinal subset, and (dually) a countable coinitial subset.
More directly:
let D be a countable dense set in L that includes the end points, if any.
Then L = bigcup{[a,b]: a,b in D; a<b} writes L as a union of countably
many closed intervals, each of which is covered by countably many O's.

I don't get it. How do you show that a closed interval is countably
covered? Why is that easier than the original problem? What am I
missing? 

Back to top 


William Elliot science forum Guru
Joined: 24 Mar 2005
Posts: 1906

Posted: Mon Aug 01, 2005 5:19 am Post subject:
Re: Separable Linear Order



From: Fred Galvin <fred.galvin@gmail.com>
Newsgroups: sci.math.research
Subject: Re: Separable Linear Order
fred.galvin@gmail.com wrote:
Quote:  William Elliot wrote:
Given a linear order L with the usual interval topology,
how does one show when L is separable, L is Lindelof?
Let C be any open cover of L; we may assume that the members of C are
open intervals (a, b). Define an equivalence relation on L so that,
when x < y, the points x and y are equivalent just in case the closed
interval [x, y] is covered by countably many members of C. Each
equivalence class is open, since it is the union of some subcollection
of C. So the equivalence classes form a collection of disjoint nonempty
open subsets of L. Since L is separable, it follows that there are only
countably many equivalence classes. To finish the proof, we have to
show that each equivalence class is covered by countably many members
of C.
From the separability of L, it follows that L cannot have a subset of
order type omega_1. Hence each equivalence class must have a countable
cofinal subset, and (dually) a countable coinitial subset. Hence each
equivalence class is the union of countably many closed intervals. By
the definition of the equivalence relation, each closed interval
(contained in an equivalence class) is covered by countably many
members of C. That does it, I think.
PS. Note that, instead of assuming that L is separable, you can get
by with the weaker assumption that L satisfies the "countable chain
condition", i.e., every collection of disjoint open sets is
countable.

Yes
separable ==> ccc (countable chain condition)
and your assertion is correct because
linear order topologies are monotonically normal and
ccc, monotonically normal ==> (hereditarily) Lindelof
A space (S,tau) is monotonically normal (MN) when
for all x, open U nhood x, some open mu(x,U) nhood x
with for all x,y, open U,V,
x in U, y in V, nonnul mu(x,U) /\ mu(y,V) ==> x in V or y in U
The partial map mu:Sxtau > tau is called a normality (for S).
Monotonically normal is hereditary and implies completely normal.
Metric spaces are MN.
I have tried to adapt your proof of
separable linear order topology ==> Lindelof
or
ccc, linear order topology ==> Lindelof
to
separable, monotonically normal ==> Lindelof
with an eye to
ccc, monotonically normal ==> Lindelof.
But I have netted a bankruptcy of ideas, unable to find any
remotely eligible equivalence relation for the first step.
Incidentally, a discrete subspace with cardinality kappa of a
monotonically normal space will yield a collection of pairwise disjoint
open sets with cardinality kappa. Thus for MN spaces, ccc and every
discrete subspace is countable, are equivalent.
 

Back to top 


alain verghote science forum Guru Wannabe
Joined: 29 Apr 2005
Posts: 293

Posted: Wed Aug 17, 2005 4:37 pm Post subject:
Re: " Polynomials and flows built on x^2 2 ... just iterates ."



Dear J Silverman,
I've read your reply about commuting polynomials
on sci.math ;but will you agree with the 'rule that
if two real polynomials p,q commute then there is always
on C a function f(x) such as :
p(x)=f(x)^[a] q(x)=f(x)^[b]
[a] and [b] real or complex iteration numbers ,
for 'coherence f(x)^[0] =x ,all (x functions)^[0]=x
Alain. 

Back to top 


JoeS science forum beginner
Joined: 13 May 2005
Posts: 37

Posted: Thu Aug 18, 2005 10:05 am Post subject:
Re: " Polynomials and flows built on x^2 2 ... just iterates ."



Hi Alain,
First, a correction. Ritt's theorem as I stated it is correct _unless_
p(x) and q(x) are both iterates of some other polynomial, i.e., I
forgot to include the trivial case where polynomials can commute.
On rereading your post, I see that I may have misunderstood. You're not
requiring that f(x) be a polynomial.
However, you are assuming that p(x) and q(x) are commuting polynomials
(let's assume of degree at least 2) with real coefficients. By Ritt's
theorem, that means that after a linear change of variables, there are
only three cases:
Case 1. p(x)=f(x)^[a] and q(x)=f(x)^[b] for some polynomial f(x) and
positive integers a and b.
This is the trivial case.
Case 2. p(x)=x^m and q(x)=x^n
So what will you take f(x) to be? I guess take f(x)=x^2, then
f(x)^[a]=x^{2^a}, so is that what you mean by allowing "real" iteration
numbers, you'll let a=log(m)/log(2), etc?
Case 3: p(x)=T_m(x) and q(x)=T_n(x) are Tchebychev polynomials
I guess you can probably do something similar here with "real"
iteration numbers, since T_m(x) is just x^m on the quotient of
projective space by the involution x > 1/x.
Cheers, Joe 

Back to top 


G. A. Edgar science forum Guru
Joined: 29 Apr 2005
Posts: 470

Posted: Thu Aug 18, 2005 12:45 pm Post subject:
Re: " Polynomials and flows built on x^2 2 ... just iterates ."



Here is the theorem of Ritt...
http://www.math.ohiostate.edu/~edgar/ritt_poly/
Presumably Mr Verghote requires the "complex" iteration numbers to take
care of the roots of unity that appear in the formula.
Perhaps he should explain that more fully.
Here is perhaps the simplest example:
p(z) = z^3+z, q(z) = z^3z.
Then p(q(z)) = q(p(z)).
How do you realize these as common iterates?

G. A. Edgar http://www.math.ohiostate.edu/~edgar/ 

Back to top 


alain verghote science forum Guru Wannabe
Joined: 29 Apr 2005
Posts: 293

Posted: Thu Aug 18, 2005 6:12 pm Post subject:
Re: " Polynomials and flows built on x^2 2 ... just iterates ."



Dear Friends,
Owing to you I may progress in my *Domaine.
> to Joe:yes I need real(or complex) iteration
numbers ( convention p^[o](x) =x ) ,
> to G.A Edgar look at this simple example:
p(x)=2x+1 q(x) =  2x  3 ,
p(q(x)) = q(p(x)) =  4x  5 ,
p, q are different roots of (4x+3)^[1/2]
Grand merci, Alain. 

Back to top 


sasha mal science forum beginner
Joined: 05 May 2005
Posts: 21

Posted: Mon Aug 29, 2005 3:30 pm Post subject:
Re: metrics on contextfree languages



Thank you very much for giving your approximation schemes. Some questions:
You say:
[
SCHEME #1: Characterization by Subwords
Any regular language can be characterized by the set of nglyphs (for n
= 0, 1, 2, ...) which it does not contain as any subword. A minimal
set is always finite. Cutting it off at n = N gives you a family of
approximating regular languages R_0, R_1, R_2, ..., that converges to
the regular language R.
An example: the language R = (ab)* over the alphabet {a,b,c} has the
minimal set {c,aa,bb}.
The characterization is unique, if you also add in extra markers for
the word boundaries (^ for beginning, $ for end), in which case the set
above is modified to { $$, ^^, $^, $a, $b, $c, a^, b^, c^ }, taken in
union with the set { c, aa, bb, ^b, a$ }.
The same is true for contextfree languages  except the sequence is
infinite, but the languages R_0, R_1, R_2, ... are still all regular.
]
What exactly ("the same") is true for contextfree languages? That the
characrerization is unique? Is this not trivial for all languages, since
the set of all nglyphs with ^ and $ contains also all words which
don't belong to the language. And the sets {^w$  w not in L} are
different for different L. Or that the languages of the sequence are
regular? This is also true, since they are finite.
Or that the minimal set (without ^ and $) is finite in the contextfree
case? I don't see that. Could you give a reference?
In SCHEME #2, you allow introduction of a star on the right side.
Although one can do it automatically in some cases, i.e. a rule
"A >= term1 + (term2) A" is transformed to "A >= term2* term1"
and
"A >= term1 + A (term2)" is transformed to "A >= term1 term2*"
but I don't see that it introduces all possibles stars. For instance,
"A >= a + aAa" gives the same language as "A >= a(aa)*". Somehow one
needs a notion of a sufficient condition on the occurences of star.
Without star, the second sceme seems to generate languages of words of
bigger and bigger length. Besides, I don't need convergence from below 
it is a simple as generating the words of L of bigger and bigger length.
With star, it improves  but still converges from below.
A simple
SCHEME #5
would give the languages
R_k := ( words_{<k}(L) union prefixes_k(L) )
of words of length at most k1 of L and prefixes of length =k of words of L.
Your SCHME #3 is good, but one has to decide whether L = L' for some
contextfree languages. I agree with you that one can reduce the general
equivalnce problem to this. By the way, in your example, you probably meant
c\N(n) = S(n+1) (and not =S(n), as you wrote)
since
c\N(n) = c \ N P* (V P*)^n = S P* (V P*)^n = N V P* P* (V P*)^n =
N V P* (V P*)^n = N (V P*)^{n+1} = S(n+1).
Then , you wrote
[
The automaton above, though infinite, is the minimal deterministic
automaton...
]
What do you mean by minimal in the infinite case?
By the way, where do you have SCHEME #3 from?
In SCHME#4, what you define is an automaton that forgets everything that
is deeper in the stack than n.
Thank you for all that.
My intention was to approximate from above and to get the language L as
an element of the approximation sequence if it turns out to be regular
by chance. Although the corresponding decision problem is undecidable,
I'm completely happy with the fact that the next element of the sequence
(after another reresenation of L) would either give the same language L
or would never be computed.
Regards,
sasha. 

Back to top 


John Baez science forum Guru Wannabe
Joined: 01 May 2005
Posts: 220

Posted: Wed Sep 21, 2005 9:35 am Post subject:
Re: This Week's Finds in Mathematical Physics (Week 221)



Some corrections and addenda:
In article <dgli7r$fb9$1@glue.ucr.edu>,
John Baez <baez@math.removethis.ucr.andthis.edu> wrote:
Quote:  The eleventh century was the golden age of Andalusian astronomy
and mathematics, with a lot of innovation in astrolabes. During
the Caliphate (9121031),

Actually the Caliphate began in 929 when Abd alRahman declared
himself caliph, though he assumed power in 912.
Quote:  three quarters of all mathematical
manuscripts were produced in Cordoba, most of the rest in
Sevilla, and only a few in Granada in Toledo.

Of course that should be "Granada and Toledo".
In response to this:
Quote:  So, medieval Europe learned a lot of Greek science by reading Latin
translations of Arab translations of Syriac translations of
secondhand copies of the original Greek texts!

a friend of mine wrote:
 This all seems so precarious a process that it makes me wonder whether
 there was ten times as much valuable ancient math and philosophy as we
 know about, most of which got *completely* lost.
Something like this almost certainly true.
Like Plato, Aristotle is believed to have written dialogs which presented
his ideas in a polished form. They were all lost. His extant writings
are just "lecture notes" for courses he taught!
Euripides wrote at least 75 plays, of which only 19 survive in their
full form. We have fragments or excerpts of some more. This isn't
philosophy or math, but it's still incredibly tragic (pardon the pun).
The mathematician Apollonius wrote a book on "Tangencies" which is lost.
Only four of his eight books on "Conics" survive in Greek. Luckily, the
first seven survive in Arabic.
The burning of the library of Alexandria is partially to blame for
these losses.
There's some good news, though:
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 Xray beam!
http://www.mlahanas.de/Greeks/ArchimedesPal.htm
http://newsservice.stanford.edu/news/2005/may25/archimedes052505.html
This manuscript also reveals for the first time that he did work on
combinatorics:
http://www.mlahanas.de/Greeks/ArchimedesComb.htm
A team using multispectral imaging has recently been able to read
parts of a Roman library that was "roasted in place"  heavily carbonized 
during the eruption of Vesuvius that destroyed Pompeii in AD 79. By
distinguishing between different shades of black, they were able to
reconstruct an entire book "On Piety" by one Philodemus:
http://magazine.byu.edu/article.tpl?num=44Spr01
The same team is now studying over 400,000 fragments of papyrus found
in an ancient garbage dump in the old Egyptian town of Oxyrhynchus. They've
pieced together new fragments of plays by Euripides, Sophocles and Menander,
lost lines from the poets Sappho, Hesiod, and Archilocus, and most of
a book by Hesiod:
http://www.papyrology.ox.ac.uk/multi/procedure.html
If you just want to look at a nice "before and after" movie of what
multispectral imaging can do, try this link.
George Baloglu recommends the following book:
Dimitri Gutas, Greek Thought, Arabic Culture: The GraecoArabic
Translation Movement in Baghdad and Early 'Abbasid Society
(2nd4th/8th10th Centuries).
Finally:
In article <dgnka4$get$1@dizzy.math.ohiostate.edu>,
Noam Elkies <elkies@math.harvard.edu> wrote:
Quote:  Amusingly, Arabic numerals were also called "dust numerals" since
they were used in calculations on an easily erasable "dust board".
Their use was described in the Liber Pulveris, or "book of dust".
This is even more amusing than you may realize: the word "abacus"
comes from a Greek word "abax, abak" for "counting board", which
conjecturally might come from the Hebrew word (or a cognate word
in another semitic language) for "dust"! See for instance
http://education.yahoo.com/reference/dictionary/entry/abacus>.
So these "dust numerals" replaced a reckoning device whose name
may also originate with calculation a dust board...

Interesting! While "calculus" refers back to pebbles. 

Back to top 


Google


Back to top 



The time now is Fri Mar 22, 2019 7:09 pm  All times are GMT

