mm-381
===
Subject:
: apologies, and new question, abstract
algebraI misread the question, missing the hypothesis that N
is a subgroup ofH...sorry about that.now, i have a new
question...if |G| = p^n where p is a prime then a subgroup H
of G s.t.|H|=p^(n-1) is normal.The suggestion is to assume H
is not normal, ie, xHx^-1 =/= H for somex in G.and to use|HK|
= |H||K|/|H intersection K| somehow to produce a
contradiction. I do not see how to do this
immediately...certainly x is not in H, and |x| = p^i for some
i>=1, and |Hintersection <x>| = p^j for some j, i > j >= 0
.however, this does not appear to cause any trouble...thanks
for the assistance.===
===
Subject:
: Re: apologies, and new
question, abstract algebra>I misread the question, missing the
hypothesis that N is a subgroup of>H...sorry about that.>now, i
have a new question...>if |G| = p^n where p is a prime then a
subgroup H of G s.t.>|H|=p^(n-1) is normal.>The suggestion is
to assume H is not normal, ie, xHx^-1 =/= H for some>x in
G.>and to use>|HK| = |H||K|/|H intersection K| somehow to
produce a contradiction. >I do not see how to do this
immediately...>certainly x is not in H, and |x| = p^i for some
i>=1, and |H>intersection <x>| = p^j for some j, i > j >= 0
.>however, this does not appear to cause any trouble...>thanks
for the assistance.Let K = xHx^-1. If H != K, then |KH| is a
power of p larger than p^(n-1),so KH = G. Hence x is in KH =
xHx^-1H implies x in H, contradiction.But personally I prefer
to deduce this result as a consequence of the factthat any
proper subgroup of a finite p-group is properly contained in
itsnormalizer, and you can prove that by induction on |G|,
using the factthat Z(G) != 1 for G nontrivial finite
p-group.Derek Holt.===
===
Subject:
: Re: Simple LOGIC Is Too
Advanced For Mainstream Economists> Would you call a biologist
an expert in his field if he didn't even know> what is
designated by the acronym DNA? How about chemist who couldn't>
explain stoichiometry? A mathematician who couldn't prove the
Pythagorean> theorem? A physicist who couldn't name Newton's
Second Law of Motion? Just about everyone in their respective
fields would call them outright> frauds or just plain stupid.
What am I saying? Most educated people> outside their field
would think there was something wrong. That's because> those
questions are so basic you can go to any college or university
and 98% of> the high schools in the U. S. and you know you'll
get the correct answer in> less than 20 seconds. The
suggestion that there would be any stonewalling is ludicrous.
Now, let's leave the reputable science and math departments at
every last> college and university on the planet and head on
over the outspoken free market> scholars at the Chicago
School, von Mises Inst., Hoover Inst., American> Enterprise,
Cato, etc. and ask them a question that is even more basic to
their> field: Does free speech precede each and every free
trade? No a war precedes each and every free trade. Since
econo weirdos are still reminded that only Euro-weirdos have
free trade. But the US has NAFTA. Even though the correct
answer is an obvious self evident truth, the outspoken> market
economists won't have any answers. In fact, these outspoken>
scholars will stonewall and dodge like Labor Secretary Chao at
a press> conference. As Milton Friedman might reason, if
corporate interests pay economists to> dodge issues, next
thing you know, you have a lot of economists who dodge>
issues. But since Milton Friedman dodges the issue of U.S.
Social Security at each and every one of his press
conferences, that's obviously the reason why the U.S. has
temporarily traded him to the U.N. and the World Bank. Since
both are well-known to be filled with Middle East
heads.===
===
Subject:
: Re: theoreme de Rademacher> .... >
d.8emonstration(s?) en fran.8dais .... Connais-tu les
newsgroups fr.education.entraide.maths et fr.sci.maths ?
===
===
Subject:
: Re: theoreme de Rademacher>.8enonc.8e du
th.8eor.8fme : une fonction lipschitzienne sur un ouvert est
presque partout>diferrentiable sur cet
ouvert.>d.8emonstration(s?) en fran.8dais, Maybe someone can
translate: In one dimension it's clearthat a lipschitz
function is (locally) absolutely
continuous.>ouvrages,...>Merci.===
===
Subject:
: Re: theoreme de
Rademacher>.8enonc.8e du th.8eor.8fme : une fonction
lipschitzienne sur un ouvert est presque
partout>diferrentiable sur cet ouvert.d.8emonstration(s?) en
fran.8dais, Maybe someone can translate.... A Lipschitz
function on an open set is differentiable almost everywhere on
that set. (The OP asked for a proof in French.) ===
===
Subject:
: Re:
theoreme de Rademacher>>>.8enonc.8e du th.8eor.8fme : >>une
fonction lipschitzienne sur un ouvert est presque
partout>>diferrentiable sur cet ouvert.>>d.8emonstration(s?)
en fran.8dais, >>> Maybe someone can translate....> A
Lipschitz function on an open set is differentiable almost
>everywhere on that set. (The OP asked for a proof in
French.)was hoping someone could translate my proof into
French.(My French is read-only, and just barely that.)>
===
===
Subject:
: Re: theoreme de Rademacher>was hoping someone
could translate my proof into French.>(My French is read-only,
and just barely that.)Which brings to mind, in how many
languages can people here read andunderstand mathematical
material? It seems like with a goodmathematical dictionary and
a grasp of the concepts behind the textone could quickly
understand most papers even when not very familiarwith the
actual language. At least as much as the paper itself
allows.Any professional mathematician I would expect to read
papers in atleast English, French, German, Latin and maybe
Spanish, Russian andItalian.-- I'm not interested in
mathematics that might have anythingto do with reality. --
Easterly, in sci.math===
===
Subject:
: Re: theoreme de
Rademacher>>was hoping someone could translate my proof into
French.>>(My French is read-only, and just barely that.)>Which
brings to mind, in how many languages can people here read
and>understand mathematical material? It seems like with a
good>mathematical dictionary and a grasp of the concepts
behind the text>one could quickly understand most papers even
when not very familiar>with the actual language. At least as
much as the paper itself allows.>Any professional
mathematician I would expect to read papers in at>least
English, French, German, Latin and maybe Spanish, Russian
and>Italian.Most of those are pretty standard, and not to hard
to read math inwith a minimal command of the language. Russian
would be _very_useful, but it's not as easy (or maybe that's
just my impression.)Luckily there's so much important Russian
math that a lot ofit gets translated for dummies like me.But
saying any professional mathematician should be able toread
math in Latin seems just silly, sorry. That would have
beenclear a few centuries ago...===
===
Subject:
: probability
questionhello,i have been trying to solve this on and off for
a while now .... cant seemto figure it out.X_i iid ~ Exp(1)S_n
= mean of X_1 ... X_n (i.e. (sum from i=1 to n of X_i)/n)Z_n =
max{S_1, S_2, ... S_n}show E(Z_n) = sum from i=1 to n of
1/(i^2)any hints???thanks!rizwan===
===
Subject:
: Re: Big Bertha
Thing blogsBig Bertha Thing progressCosmic Ray SeriesPossible
Real World System
Constructshttp://web.onetel.com/~tonylance/progress.htmlAccess
page to 6K Web page Astrophysics net ring access siteNewsgroup
Reviews including uk.transportThe Progress of Discontentby T.
Warton(Written at Oxford in 1746) Half-hours With the Best
AuthorsThe Chandos ClassicsEdited by Charles KnightPublished
by Frederick Warne and Co. 1890 Big Bertha Thing besidesSome
people think that Big Bertha is less suitable for the Have
they heard the long name for CP Conf? Perspective. They think
that is a better place to put Big Bertha.They must be besides
themselves, which proves the thing. (C) Copyright Tony Lance
1998To comply with my copyright,please distribute complete and
free of charge. Tony Lancejudemarie@big-bertha-thing.comBig
Bertha Thing debitUK big banks have now made debit cards
mandatory.1. Every cheque book has a debit card.2. You do not
sign agreement for debit card.3. Lost or stolen cards must be
cancelled.4. Insure debit cards for 1000 pounds sterling.5.
For every 8 pounds loss charged to the customer, the bank can
lend 92 pounds. (Multiplier)6. Cheque guarantee card
included.7. Cash Card included.8. Phantom withdrawals
included.9. 50 pounds cashback at supermarkets.10. 250 pounds
'Hole-in-the-wall' withdrawals.11. 100 pounds Post Office and
bank withdrawals.12. How do you prove that you have cut the
card up and disposed of it?13. How do you prove that you did
not use it before destruction?14. Cut the card up and give to
your solicitor or third party for safe-keeping. (Escrow)15.
The survival of the banks depends on putting customers to the
sword. (Barbaric) Obviously not all customers, just enough.16.
Unauthorized access to debit card accounts, includes both
criminals and bank insiders, which seem to be synonymous.17.
Authorized access to debit card accounts, includes bank
insiders, grazing on the customers like milk cows, in their
official capacity, as per job description.18. You no longer
need to prove this in a court of law, since it is endemic and
ubiquitous. (Common knowledge) Case law sufficient for class
action.===
===
Subject:
: Re: Standards of Measure, what were they
based on? was Re: Geodetic Measurements of the Great Pyramid
aka 1/43,200 by support1.mathforum.org (8.11.6/8.11.6/The Math
Forum, $Revision: 1.9 primary) id i2F3I0a20067;><pre[many
snips]>>says...>I asked you how you could tell that a ratio
was part of the design>rather than coincidental. You did not
answer. I ask if all ratios were>important, you responded
other than no. What I really want to know,>because the
solution would be so useful, is how you can tell if the>ratio
is part of the design.>>A design implies a plan. A plan
implies intent.>Really? A beehive has a rather nice design
that appears not to be the>result of plan or intent.>> If the
presence of>>a ratio is an intentional part of a design, it
might have as its>>purpose a desire to organise, or control,
or modulate the arrangement>>of the work in some harmonious
arithmetic or geometric progression.>This is a completely
non-responsive reply. Or in other words, you>still haven't
answered the question. We don't want to know what the>purpose
of an intentionally incorporated ratio might be; we want
to>know how to tell when the ratio's presence is intentional.
Not when>it might be intentional, mind you, but when it
actually *is*.>Admittedly, there are phrases in your claim
that have no >meaning that I can discern (measured
architectural>modulor, certain natural osophy).>>Read any good
book on architecture; Vitruvious, through
Ching.>Vitruvius.>>1.) measured architectural modulor;
(Corbusier)>> A.) Is there a difference between making a
single point>> and offering another point which leads to a
line of thought,>> an area of discussion and a body of
knowledge?>Explain yourself. Are these geometrical points,
debating points,>points of the compass, or what? And what have
they to do with>measured architectural modulors, which you
still haven't bothered to>define?>> B.) Can the resolution of
opposites lead to a progressive>> series of transformations
between form and space? (Ching)>And in real English this
means? And bears what relation to measured>architectural
modulors?>> C.)See also; Ordering Principles; Axis, Symmetry,
>> central, linear, radial, cluster and grid; systems>> of
proportion, the golden section, greek orders,>> renaissance
theories, modulor and ken.>As mathematicians we know a lot
about ordering principles, axes,>symmetry of various kinds,
etc. Modulors are beyond our ken.>Apparently you'd rather
mystify than enlighten, since you still>haven't said anything
useful on the subject.>> D.) Ditical process; standards;
paired opposites; >> similarity, difference, motion, rest,
number, sequence, consequence.>Yes, we are familiar with these
concepts. Can you use them to define>measured architectural
modulors?>> F.) Formal collisions of geometry; separations and
links;>> rotated grids; edges and corners. >And a formal
collision of geometry is? Separations and links of what?>>2.)
natural osophy; try Vitruvious; Elements; >> Air; Earth; Fire;
Water; (Mesopotamian origin)>> Properties; Hot, Cold, Wet,
Dry.>I don't think so. Ancient natural osophy contains
some>intriguingly elaborate intellectual systems, but I prefer
modern>chemistry and physics.>???? Please explain this in a
little more detail.>>The Japanese used the Ken See
Architecture, Form, Space and Order>>Francis DK Ching, Van
Nostrand Reinhold, 1979, p 320>>The traditional Japanese unit
of measure the Shaku was >>originally imported from China and
is almost equivalent to >>the English, (Greek, Roman and
Egyptian) foot.>Pedal extremities being much of a muchness in
most parts of the world,>I can't say that I'm overwhelmed by
this news.>>The Ken was imported during Japans Middle
Ages.>>The Ken mat series is 3, 4, 4 1/2, 6, 8, 10. >This is
meaningless without more context.>>In the Kyo-mo method the
floor mat remained constant>>in a ration of Pi to 2 Pi Shaku
and the column spacing or Ken>>varied from 6.4 to 6.7 Shaku.
The ceiling height is determined by the>>number of mats x
.03>The ratio of pi to 2 pi is 1/2; what has pi to do with
it?>> But, better yet,>please answer my questions
above.>>ok>And why have you given>different information than I
asked? Ratios should be unit free.>>The Egyptians didn't use
ratios, they used unit fractions.>>That makes the units
important.>PI is a ratio. Unit fractions, like decimal
notation, are a way of>describing ratios. As such they are
both unit free. Now why don't you>actually answer the question
asked?>>See the quotes from Vitruvious...>They didn't answer
any of the questions.>>If you want me to validate your
assertion; even though>>is based on assumptions which I don't
share, we are going to>>have to find some common ground. >>1.)
The Egyptians didn't use ratios>>2.) The did use unit
fractions>>3.) Their units were generally standard units of
measure>>4.) Although such modulors are common in architecture
today>>they were not so common c 3000 BC. >>Perhaps you might
address yourself to those points...>1 and 2) Unit fractions
are a different notation from decimal, but do>not say anything
essentially different. I would write 3.142 as 3/1 +>1/10 +
4/100 + 2/1000. They make the same statement. Unit
fractions>are ratios. >>That is a series of fractions; not
unit fractions.>So? He could, with a bit more work, have
written 3.142 as 3 + 1/8 +>1/60 + 1/3000, which *is* a
representation in terms of unit fractions.>The point is that
3.142, his representation, mine, 1571/500, etc. are>all
representations of the same number. The Egyptians used
an>uncommonly clumsy system for writing fractions, but that is
a>superficiality: the numerical facts aren't changed by the
notation.>Oh, and unit fractions are *still* ratios.>3)
Generally standard I don't know what that means. But I doubt
it>means that they would change the units in mid
construction.>>The best reference to standards would be the
Platonic Dialogs>>which explains very well the Egyptian
conmcept of Ma'at.>That, sir, is a fake citation. What
dialogue? Where? And how is>this concept involved in changing
units in mid-construction?>4) I have no idea what a modulor
is.>>Its an architectural system of modules related to the
Fibonacci series.>And a module in this context is? How is it
related to the Fibonacci>sequence?>Brian M.
Scott></pre>===
===
Subject:
: Royden qn by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
i2F3I1F20156;vector space = VSfunctions = fcns Need to show 1)
X is a topological VS which does not admit any cont. linear
functionals.2) X is not locally convexwhere X is the VS of all
(real) valued fcns on [0,1 ] with + & * defined in the usual
way. X has metric topology defined by int_0^1 frac{left|
x(s)-y(s)right|}{1+left|x(s)-y(s)right|} ,ds I think this qn
is similar to Royden's #49 pg 244. In that qn I can prove (a),
(c), (d) & (e) but I am still having troubles with the parts
that are like 1) & 2) above.===
===
Subject:
: Re: Royden qn>vector
space = VS>functions = fcns>Need to show >1) X is a
topological VS which does not admit any cont. linear
functionals.>2) X is not locally convex>where X is the VS of
all (real) valued fcns on [0,1 ] with + & * defined in the
usual way. X has metric topology defined by >int_0^1
frac{left| x(s)-y(s)right|}{1+left|x(s)-y(s)right|} ,ds You
really think that's easier to read than int_0^1 |x(s)-y(s)| /
(1+ |x(s)-y(s)|) ds ??? Anyway, this integral does not exist
for all x, y in your vectorspace. How does the problem
actually read?>I think this qn is similar to Royden's #49 pg
244. In that qn I can prove (a), (c), (d) & (e) but I am still
having troubles with the parts that are like 1) & 2)
above.===
===
Subject:
: Re: basic query by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
i2F3I3x20211;>Hi I am currently an A-level maths student in
Englnd I am 17 years old>and just need a little basic help
with my understanding of>differentiating an equation wih a
square using the double>differentiating from first principles
formula thanx.>Also can someone explain to me the basic
conditional probability>formula for a intersection b and also
a union b. Thanx alot. According to Your probability
questions, I think the bestillustrating example is once using
two groups of people,let two classes of pupils: A with A
pupils and B with B pupils; let in the class A the number of
girls is aand in the class B the number of girls is b; Now the
probability, that You meet one girl member of two thisclases is
P = (a+b)/(A+B) once some pure situation:what illustrates
simple union of a and b probabilities; but Youneed imagine,
that this two classes can contain mixed members:lets take,
that 5 girls from class-course A are also in the class-course
B . Now union chance of a+b to meet some girl will be
different: P = (a+b-5-5)/(A+B-5-5) For to show intersection it
will be just to ask for the chanceto meet the girl, wchich is
attending the class A and B simultanously: P = 5/(A+B-5-5)
Once also we need to take, that no one same boy is
attendingthese two classes (courses). For small sample
classes: A=7;a=3;B=10;b=2 and one girl attending to both of
them: P(a union b) = 3/15 = 1/3 P(a intersection b) = 1/15
There are some grafical illustrations of such propertiesand
You can find it in several literatures: for example: Kai Lai
Chung - Elementary Probability Theory... Ro ===
===
Subject:
: Happy
Pi DayI rarely post on the news group, but I read it
alot.However I just wanted to wish everyone a Happy Pi
Day.March 14th3/14Also Einstein's birthday.===
===
Subject:
: Re:
Happy Pi Day> I rarely post on the news group, but I read it
alot.> However I just wanted to wish everyone a Happy Pi
Day.March 14th> 3/14> Of course in Europe, Pi Day is
celebrated on 31 April: 31/4.-- G. A. Edgar
http://www.math.ohio-state.edu/~edgar/===
===
Subject:
: Re: Happy Pi
DayI rarely post on the news group, but I read it alot.>
However I just wanted to wish everyone a Happy Pi Day.March
14th> 3/14Of course in Europe, Pi Day is celebrated on 31
April: 31/4.You may have meant this humorously anyway, but I
believe that even inEurope there are only 30 days in
April.:)But pi IS irrational,...Leroy Quet===
===
Subject:
: Re:
Happy Pi DayX-ID:
Z7zBh-ZV8egO+0BMjSv2W+12VJf7beQHgtPz7hf1EgtaBtEYKv8Oo1 G. A.
Edgar <edgar@math.ohio-state.edu.invalid> schrieb:>> I rarely
post on the news group, but I read it alot.>> However I just
wanted to wish everyone a Happy Pi Day.>>> March 14th>>
3/14>>Of course in Europe, Pi Day is celebrated on 31 April:
31/4.Yup, that's actually a holiday in Germany. It's called Tag
derArbeit (labour day).===
===
Subject:
: Re: Happy Pi Day> G. A.
Edgar <edgar@math.ohio-state.edu.invalid> schrieb:> I rarely
post on the news group, but I read it alot.> However I just
wanted to wish everyone a Happy Pi Day.March 14th> 3/14>>Of
course in Europe, Pi Day is celebrated on 31 April: 31/4.>
Yup, that's actually a holiday in Germany. It's called Tag
der> Arbeit (labour day).So, to paraphrase Lewis Carroll, we
can say that on 30 April there is Pitomorrow, and on 1 May
there is Pi yesterday, but there is never Pitoday.-- Dave
SeamanJudge Yohn's mistakes revealed in Mumia Abu-Jamal
ruling.<http://www.commoncouragepress.com/index.cfm?action=
book&bookid=228>===
===
Subject:
: Re: Happy Pi Day> G. A. Edgar
<edgar@math.ohio-state.edu.invalid> schrieb:> I rarely post on
the news group, but I read it alot.> However I just wanted to
wish everyone a Happy Pi Day. March 14th> 3/14>Of course in
Europe, Pi Day is celebrated on 31 April: 31/4.> Yup, that's
actually a holiday in Germany. It's called Tag der> Arbeit
(labour day).> So, to paraphrase Lewis Carroll, we can say
that on 30 April there is Pi> tomorrow, and on 1 May there is
Pi yesterday, but there is never Pi> today.There will be pi in
the sky when you die.--W. Guthrie.===
===
Subject:
: Re: Happy Pi Day>
I rarely post on the news group, but I read it alot.> However I
just wanted to wish everyone a Happy Pi Day.> March 14th> 3/14I
think 22 July is closer.===
===
Subject:
: Re: Happy Pi Day> I rarely
post on the news group, but I read it alot.> However I just
wanted to wish everyone a Happy Pi Day. March 14th> 3/14I
think 22 July is closer.> 3/14 1:59:26March 14, 1:59 o'clock
and 26 seconds===
===
Subject:
: There needs to be ANSI or ISO math
standardsThere needs to be ANSI or ISO math standardssome
teachers are very anal about symbols and other stuff on tests
and homeworkand i do like like being out floating on air, i
rather have some ground below my feet on things likeabs(x-4) <
E and not have a teacher bitch about me not having E such that
E > 0 before that statement even though the abs states it
implicitlyor if i do not have an = sign for 4+x =51when the =1
the equal sign is implicit again.please tell me there is a math
ANSI or ISO because math is old and teachers can be
jerks===
===
Subject:
: Re: There needs to be ANSI or ISO math
standardsI thought, it was one of his items, but, anyway,This
is a job for Jimi Harris and the Math-sub-nu Experience!
anywaay, hey're not anal about, except insofaras rectalinear.>
There needs to be ANSI or ISO math standardssome teachers are
very anal about symbols and other stuff on tests and >
homework as for Science Friday, that's just one of the reasons
that I stoppedlistening to the God-am radio, as with the Boob
Tube, before that;NPR's main job, now, is to assure thattwo
Bonesman vault out of a mausoleum on the Yale campus,and
hokey-dokey Copenhagen Schooler crappola just addsto the smoke
(note tahtUsama bin Laden is in a perfectly undead
superposition,with reference to the Bush and Media Hype,
althoughit wouldn't matter weather this state collapsedto dead
or alive, to Blair's McCrusade ... in the venueof
Schroedinger's joke about the Cat in the Boxwith the Atom to
Its Head .-) thus quoth:just doesn't make sense., even to the
phyicists! It seems that as we approach the limits of our
current understanding of themicro and macrocosms, many of our
concepts begin to break down. Buckycontended unflinchingly
that the problem was our use of Cartesian --Give Earth a
Trickier Dick Cheeny -- out of office, after GIGA
years.http://www.benfranklinbooks.com/http://www.rand.org/
publications/randreview/issues/rr.12.00/http://
members.tripod.com/~american_almanac===
===
Subject:
: Re: There
needs to be ANSI or ISO math standards>There needs to be ANSI
or ISO math standards>some teachers are very anal about
symbols and other stuff on tests and >homework>and i do like
like being out floating on air, i rather have some ground
>below my feet on things like>abs(x-4) < E >and not have a
teacher bitch about >me not having >E such that E > 0 >before
that statement even though the abs states it implicitly>or if
i do not have an = sign for >4+x =5>1>when the >=1 the equal
sign is implicit again.So you're saying it's clear that you
meant to say 4+x=5=1??? HINT: _That_ says that 5 = 1. That's
really what youmeant to say, 5 = 1?>please tell me there is a
math ANSI or ISO because math is old and teachers >can be
jerks===
===
Subject:
: Re: There needs to be ANSI or ISO math
standardsMathematics must be written so that it is impossible
to misunderstand,not merely so that it is possible to
understand. When you learn towrite it so your teacher (even if
a jerk) can find NO POSSIBLEobjections, you will have achieved
something. Something useful in yourfuture mathematics studies.
I think your teacher is doing the rightthing.-- G. A. Edgar
http://www.math.ohio-state.edu/~edgar/===
===
Subject:
: Re: There
needs to be ANSI or ISO math standardsJohn Cho a .8ecrit:>
There needs to be ANSI or ISO math standards> It's not the
problem here> some teachers are very anal about symbols and
other stuff on tests and > homeworkand i do like like being
out floating on air, i rather have some ground > below my feet
on things likeabs(x-4) < E and not have a teacher bitch about >
me not having E such that E > 0 What you say here is floating
on air. To be standing firmly, you have to specify what is x
AND what is E. You must also specify if what you say is a
statement, an equation, etc...If x and E are specified, and
what you say is a statement then, of course e>0 is a
conseqeunce. But that means, earlier, you have explained what
is x and E (reals numbers I suppose). So if you impose E to be
a rela number or E to be a positive real number, your problem
is the same : to satisfy your teacher you have to indicate
what is E.If I was your teacher, I could say something like
:let E=+ infinity so your statement is irrelevant. I give you
0 point...> before that statement even though the abs states
it implicitly> No implicit in mathematics. Just say what you
mean, no more no less.> or if i do not have an = sign for 4+x
=5> 1> Why not x 4 5 ; 1 ?If you think mathematics is just a
collections of equations you are out of place. You have to say
what you mean. :Let x be a real number who satisfy the equation
4+x=5.We have x=1.Does it cost you anything to write correctly
?> when the =1 the equal sign is implicit again.> Implicit
statement means miscomprehension.> please tell me there is a
math ANSI or ISO because math is old and teachers > can be
jerks> No because mathematicians learn to say exactly what
they mean.And young teachers are even more evil than old ones,
I assure you...===
===
Subject:
: Re: There needs to be ANSI or ISO
math standards> There needs to be ANSI or ISO math
standardssome teachers are very anal about symbols and other
stuff on tests and > homeworkand i do like like being out
floating on air, i rather have some ground > below my feet on
things likeabs(x-4) < E and not have a teacher bitch about >
me not having E such that E > 0 > before that statement even
though the abs states it implicitlyNo it doesn'tabs(x-4) <
-1is a perfectly valid inequality. One that has no solutions,
but that's irrelevant. And you may notice that -1 is not > 0.>
or if i do not have an = sign for 4+x =5> 1when the =1 the
equal sign is implicit again.Oooooh, no it isn't. Do you
really want to claim that 5 = 1?0/2 Please try harder.-- 1st
bug in MS win2k source code found after 20 minutes:
scanline.cpp2nd and 3rd bug found after 10 more minutes:
gethost.cBoth non-exploitable. (The 2nd/3rd ones might be,
depending on the CRTL)===
===
Subject:
: Re: There needs to be ANSI
or ISO math standards> and i do like like being out floating
on air, i rather have some ground > below my feet on things
like> abs(x-4) < E > and not have a teacher bitch about me not
having > E such that E > 0 > before that statement even though
the abs states it implicitlyI'd agree that you should not have
to state that E > 0 here.> or if i do not have an = sign for >
4+x =5> 1> when the > =1> the equal sign is implicit
again.This one I'd say you are wrong on. You need to say x=1,
because otherwisethe teacher cannot be sure that you intended
1 as the answer, rather thansome intermediate value in your
failed attempt to find the answer. Yes, Iknow that 1 is the
answer, but you have to show that you know that.-- --Tim
Smith===
===
Subject:
: Re: There needs to be ANSI or ISO math
standards>> or if i do not have an = sign for>> 4+x =5>> 1>>
when the>> =1>> the equal sign is implicit again.This one I'd
say you are wrong on. You need to say x=1, because> otherwise
the teacher cannot be sure that you intended 1 as the answer,>
rather than> some intermediate value in your failed attempt to
find the answer. Yes, I> know that 1 is the answer, but you
have to show that you know that.This is the accepted practice
in writing English. Sentences must havemain verbs. The OP's
first sentence is fine: four plus 'x' equals five',but the
second sentence: one is no sentence, lacking a main verb.This
is basic literacy, not mathematics.-- Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92;
mistakes his penis for a square root, 88-9Francis Wheen, _How
Mumbo-Jumbo Conquered the World_===
===
Subject:
: Re: There needs to
be ANSI or ISO math standards> and i do like like being out
floating on air, i rather have some ground> below my feet on
things like> abs(x-4) < E> and not have a teacher bitch about
me not having> E such that E > 0It makes difference if e is
positive, negative or zero. To wit:When e <= 0, there are no
solutions.When e > 0, then 0 < x - 4 < e or 0 < -(x - 4) < eso
4 < x < e + 4 or 4 - e < x < 4are the solutions> before that
statement even though the abs states it implicitly> or if i do
not have an = sign forIt does not. For example, solve for x |x|
< -1Does the | | implicitly require -1 to be positive?No, some
inequalities don't have any solutions.> please tell me there
is a math ANSI or ISO because math is old and teachers> can be
jerksAttention to numerous small details is required skill for
mathematics.If the E to which you refer is the epsilon used in
limits, thendo note all such theorems all have the premise
epsilon > 0.Please you take my reply to your teacher and with
your preferenceof relish or ketchup, proudly eat your words in
front of her.===
===
Subject:
: Re: There needs to be ANSI or ISO math
standards<snip> please tell me there is a math ANSI or ISO
because >>math is old and teachers>> can be jerks>Attention to
numerous small details is required skill for mathematics.Also,
programming, any flavor of engineering, architecure,
farming,piloting any flavor of vehicle, bcing checkbooks.>If
the E to which you refer is the epsilon used in limits,
then>do note all such theorems all have the premise epsilon >
0.>Please you take my reply to your teacher and with your
preference>of relish or ketchup, proudly eat your words in
front of her.A little essence of crow will do./BAHSubtract a
hundred and four for e-mail.===
===
Subject:
: Re: Roots to
Polynomial Series> Sometimes it is desirable to obtain a
solution for a function> that can only be expressed as an
infinite series. This method> (see links), allows the ability
to select any number of terms> into the series according to
the precision desired, and> calculates all solutions.The
polynomial solution doesn't turn into another> polynomial
problem.http://mypeoplepc.com/members/jon8338/polynomial/http:
//www.geocities.com/jongiff2000/roots_to_any_
polynomial.htmlHow do you expand 2(2 c(3)^(2/3)
-c(2))^(1/2)into a polynomial? Newton's binomial formula will
give an infinite series. You really should read some Galois
theory. Then you'll realize whatyou're trying to do here is
impossible, and the theorem is quite explicitabout these types
of expressions. -- Christopher Heckman===
===
Subject:
: Re: Why are
there 63360 inches in a mile?Originator:
baez@math-cl-n03.math.ucr.edu (John Baez)http://www.pobonline.com/CDA/ArticleInformation/features/BNP_
_Features__Item/0,2338,88115,00.html>concludes that the perch,
defined as 16 1/2 feet (5 1/2 yards) in theAssize of weights
and measures of 1270, was already an established>measure, and
notes that it is within a fraction of an inch of 16
Danish>fods or Prussian Rheinfuss.this
subject:http://math.ucr.edu/home/baez/inches.htlm===
===
Subject:
:
Re: Why are there 63360 inches in a mile?Originator:
richard@cogsci.ed.ac.uk (Richard Tobin)Incidentally, when I
was at school (aged about 6) in the 1960s, we hadto do
arithmetic with miles, furlongs, chains, yards, feet and
inches(no rods/poles/perches, I'm afraid). You did addition in
the usualway, but instead of carrying uniformly at 10, you
carried at 12, 3,22, 10 and 8.Similarly for pounds, shillings
and pence.This meant that when we came to do number bases a
few years later, itwas trivial by comparison. No wonder
children find it harder now whenalmost everything in real life
is base 10.-- Richard===
===
Subject:
: Re: Why are there 63360 inches
in a mile?Richard Tobin wibbled:> Incidentally, when I was at
school (aged about 6) in the 1960s, we had> to do arithmetic
with miles, furlongs, chains, yards, feet and inches> (no
rods/poles/perches, I'm afraid). You did addition in the
usual> way, but instead of carrying uniformly at 10, you
carried at 12, 3,> 22, 10 and 8.Similarly for pounds,
shillings and pence.This meant that when we came to do number
bases a few years later, it> was trivial by comparison. No
wonder children find it harder now when> almost everything in
real life is base 10.-- Richard> you're me, you are===
===
Subject:
:
Abstract Algebra Question, Please Help!On a recent homework
assignment, we were given the following question:If, in a
ring, each element equals its SQUARE, show thatmultiplication
must be commutative.To which I answered:0 =
0abab-ababa-abab+ababa =
baba-baba-ababa+ababaabab-ababa-abaab+abaaba =
baba-baaba-ababa+abaaba(ab-aba)^2 = (ba-aba)^2ab-aba =
ba-abaab = baBut now we've been assigned this similar
question:If, in a ring, each element equals its CUBE, show
that multiplicationmust be commutative.And I am baffled.
Anyone got any hints?===
===
Subject:
: Re: Abstract Algebra
Question, Please Help!I received an email from Dik Winter
proclaiming that he would *not*remove my writing from his
webpage, and he claims he is not violatinginternational
copyright laws.Dik Winter is in the Netherlands, and I think
his bravado is based onconfidence that he's safe in his
country from worries about such pettythings as copyright, but
I want you all to consider his behavior incontext.For
instance, Dik Winter's argument would mean that he can take
BenPeterson's post here, and put it on his own webpage.> On a
recent homework assignment, we were given the following
question:If, in a ring, each element equals its SQUARE, show
that> multiplication must be commutative.To which I answered:0
= 0> abab-ababa-abab+ababa = baba-baba-ababa+ababa>
abab-ababa-abaab+abaaba = baba-baaba-ababa+abaaba> (ab-aba)^2
= (ba-aba)^2> ab-aba = ba-aba> ab = ba> On that webpage he
might comment on Ben Peterson's lack of grasp ofthis or that,
or say that maybe Ben Peterson just has issues which iswhy
he's stupid enough to post on Usenet, when anyone can copy
hispost to their webpage, and comment about it.> But now we've
been assigned this similar question:If, in a ring, each element
equals its CUBE, show that multiplication> must be
commutative.And I am baffled. Anyone got any hints?Or, someone
could just wait until they *knew* Ben Peterson, andpossibly was
in conflict with him or fighting for a job promotion orwho
knows what, and thought it might help to rattle him by putting
upthis post, where he's asking question--hey YEARS from now it
might berelevant--to push the case that he isn't an
independent thinker andrelies to heavily on the opinion of
others.A Dik Winter might then argue that Ben Peterson has
waived any rightsto keep his post from being put on webpages
because he posted onUsenet.And when Ben Peterson emailed him
to cry foul, he might email backfrom the Netherlands basically
laughing in his face, thinking that inthe Netherlands,
copyrights don't matter.Usenet is obviously just a place where
people can cruise for neatthings to put on their webpages,
right?James Harris===
===
Subject:
: Re: Abstract Algebra Question,
Please Help! Discussion, linux)> Usenet is obviously just a
place where people can cruise for neat> things to put on their
webpages, right?Well, that and neat things to put at the bottom
of their posts, yes.-- What I represent is the unknowable
future--the power of change. Inthat sense I'm a force of
Nature, a force of the Universe, a livingemodiment of change
itself. --James Harris and his sense of humility===
===
Subject:
:
Re: Abstract Algebra Question, Please Help!... > For instance,
Dik Winter's argument would mean that he can take Ben >
Peterson's post here, and put it on his own webpage.Perhaps.
To what purpose? Now if he was somebody who for some 8
yearshad, with high regularity, posted incorrect proofs,
defending his proofs,meanwhile insulting everybody, it would
have been something else.So Ben Peterson need not be afraid.
On the other hand, I could ofrefrain from that. Perhaps in
some 5 years when he starts incessantlyposting incorrect
proofs.-- dik t. winter, cwi, kruislaan 413, 1098 sj
amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn
amsterdam, nederland; http://www.cwi.nl/~dik/===
===
Subject:
: Re:
Abstract Algebra Question, Please Help!>I received an email
from Dik Winter proclaiming that he would *not*>remove my
writing from his webpage, and he claims he is not
violating>international copyright laws.>Dik Winter is in the
Netherlands, and I think his bravado is based on>confidence
that he's safe in his country from worries about such
petty>things as copyright, but I want you all to consider his
behavior in>context.>For instance, Dik Winter's argument would
mean that he can take Ben>Peterson's post here, and put it on
his own webpage.>> On a recent homework assignment, we were
given the following question:>>> If, in a ring, each element
equals its SQUARE, show that>> multiplication must be
commutative.>>> To which I answered:>>> 0 = 0>>
abab-ababa-abab+ababa = baba-baba-ababa+ababa>>
abab-ababa-abaab+abaaba = baba-baaba-ababa+abaaba>> (ab-aba)^2
= (ba-aba)^2>> ab-aba = ba-aba>> ab = ba>>On that webpage he
might comment on Ben Peterson's lack of grasp of>this or that,
or say that maybe Ben Peterson just has issues which is>why
he's stupid enough to post on Usenet, when anyone can copy
his>post to their webpage, and comment about it.Yes, that's
correct. What's your point in continuing to point outthings
that nobody disagrees with?>> But now we've been assigned this
similar question:>>> If, in a ring, each element equals its
CUBE, show that multiplication>> must be commutative.>>> And
I am baffled. Anyone got any hints?>Or, someone could just wait
until they *knew* Ben Peterson, and>possibly was in conflict
with him or fighting for a job promotion or>who knows what,
and thought it might help to rattle him by putting up>this
post, where he's asking question--hey YEARS from now it might
be>relevant--to push the case that he isn't an independent
thinker and>relies to heavily on the opinion of others.If you
think you can post things in public and then depend onpossible
future employers not being aware of the things you'vesaid
you're an idiot.>A Dik Winter might then argue that Ben
Peterson has waived any rights>to keep his post from being put
on webpages because he posted on>Usenet.>And when Ben Peterson
emailed him to cry foul, he might email back>from the
Netherlands basically laughing in his face, thinking that
in>the Netherlands, copyrights don't matter.>Usenet is
obviously just a place where people can cruise for neat>things
to put on their webpages, right?>James Harris===
===
Subject:
: Re:
[JSH] Abstract Algebra Question, Please Help!JSH has no
business in a thread about abstract algebra, when he cannot
even handle the ordinary kind.Besides which, his post is OT
wrt the OP.===
===
Subject:
: Re: Abstract Algebra Question, Please
Help!>On a recent homework assignment, we were given the
following question:>If, in a ring, each element equals its
SQUARE, show that>multiplication must be commutative.>To which
I answered:>0 = 0>abab-ababa-abab+ababa =
baba-baba-ababa+ababa>abab-ababa-abaab+abaaba =
baba-baaba-ababa+abaaba>(ab-aba)^2 = (ba-aba)^2>ab-aba =
ba-aba>ab = ba>But now we've been assigned this similar
question:>If, in a ring, each element equals its CUBE, show
that multiplication>must be commutative.>And I am baffled.
Anyone got any hints?This is a notoriously difficult exercise,
but it can be done by elementarymanipulations. I would
recommend that you work at it more before givingup and looking
at a solution. Derek Holt.===
===
Subject:
: Re: Abstract Algebra
Question, Please Help!
<c33t48$pd0$1@wisteria.csv.warwick.ac.ukIf, in a ring, each
element equals its SQUARE, show that>multiplication must be
commutative.If, in a ring, each element equals its CUBE, show
that multiplication>must be commutative. This is a notoriously
difficult exercise, but it can be done by elementary>
manipulations. I would recommend that you work at it more
before giving> up and looking at a solution.If is R a ring
with for all r in R, r = r^3, then 3R is a Boolean subring.2a
= (2a)^3 = 8a^3 = 8a; 6a = 0a + aa = (a + aa)^3 = a^3 + 3a^4 +
3a^5 + a^6 = a + 3aa + 3a + aa0 = 3aa + 3a; 3a = 3aa; (3a)^2 =
9aa = 3aa = 3a----===
===
Subject:
: Re: Abstract Algebra Question,
Please Help!===
===
Subject:
: Abstract Algebra Question, Please
Help! >If, in a ring, each element equals its SQUARE, show
that >multiplication must be commutative. >0 = 0
>abab-ababa-abab+ababa = baba-baba-ababa+ababa
>abab-ababa-abaab+abaaba = baba-baaba-ababa+abaaba >(ab-aba)^2
= (ba-aba)^2 >ab-aba = ba-aba >ab = ba(B,*,+) Boolean ring when
(B,*,+) ring with aa = a.Boolean ring B ==> a + a = 0, ab = ba
a+b = (a+b)(a+b) = aa + ab + ba + bb = a + ab+ba + b for all
a,b, ab+ba = 0; a + a = aa+aa = 0; ab = ba-- >If, in a ring,
each element equals its CUBE, >show that multiplication must
be commutative.ring R, for all x in R, x^3 = x ==> R
commutative, for all x, 6x = 0 (ab^2 - b^2 ab^2)^2 = ab^2 ab^2
- ab^4 ab^2 - b^2 ab^2 ab^2 + b^2 ab^4 ab^2 =0 ab^2 - b^2 ab^2
= (ab^2 - b^2 ab^2)^3 = 0 (b^2 a - b^2 ab^2)^2 = b^2 ab^2 a -
b^2 ab^2 ab^2 - b^2 ab^4 a + b^2 ab^4 ab^2= 0 b^2 a - b^2 ab^2
= (b^2 a - b^2 ab^2)^3 = 0 ab^2 = b^2 ab^2 = b^2 a xy = xyxyxy
= x yxyx y = x yyxyx = yyxxyx = yyyxxx = yx 8x = 8x^3 = (2x)^3
= 2x-- Jacobson Theoremring R, for all r in R, some n > 1 with
r^n = r ==> R commutative thus finite cancellation rings are
commutative----===
===
Subject:
: Re: Abstract Algebra Question,
Please Help!actually assigned as two parts of a single
question on weeklyassignment #3, worth 4 points. When no one
in the class got eitherquestion, he assigned them again on
assignment #4 for 5 points. Thatweek many people got the first
part (x=x^2), but no one got the secondpart (x=x^3). So he
assigned the second part again on assignment #5for 6 points.
After many hours of playing with the numbers, and manystudy
groups of 7 people, I am only now resorting to the
newsgroup.===> 
===
Subject:
: Abstract Algebra Question, Please
Help! >If, in a ring, each element equals its SQUARE, show
that>multiplication must be commutative.> >0 = 0abab-ababa-abab+ababa = baba-baba-ababa+ababaabab-ababa-abaab+abaaba = baba-baaba-ababa+abaaba(ab-aba)^2 = (ba-aba)^2>ab-aba = ba-aba>ab = ba(B,*,+)
Boolean ring when (B,*,+) ring with aa = a.> Boolean ring B
==> a + a = 0, ab = ba> a+b = (a+b)(a+b) = aa + ab + ba + bb =
a + ab+ba + b> for all a,b, ab+ba = 0; a + a = aa+aa = 0; ab =
ba-->If, in a ring, each element equals its CUBE,>show
that multiplication must be commutative.ring R, for all x in
R, x^3 = x ==> R commutative, for all x, 6x = 0> (ab^2 - b^2
ab^2)^2> = ab^2 ab^2 - ab^4 ab^2 - b^2 ab^2 ab^2 + b^2 ab^4
ab^2 => 0> ab^2 - b^2 ab^2 = (ab^2 - b^2 ab^2)^3 = 0 (b^2 a -
b^2 ab^2)^2> = b^2 ab^2 a - b^2 ab^2 ab^2 - b^2 ab^4 a + b^2
ab^4 ab^2> = 0> b^2 a - b^2 ab^2 = (b^2 a - b^2 ab^2)^3 = 0
ab^2 = b^2 ab^2 = b^2 a> xy = xyxyxy = x yxyx y = x yyxyx =
yyxxyx = yyyxxx = yx> 8x = 8x^3 = (2x)^3 = 2x-- Jacobson
Theorem> ring R, for all r in R, some n > 1 with r^n = r ==> R
commutative> thus finite cancellation rings are
commutative----===
===
Subject:
: Multigrid MethodDear all,I got a
problem on multigrid method. I use Algebraic Multigrid Method
tosolveA x = b, where A is the following matrix[ 2 -1 -1 2 -1
-1 2 -1.... ........ -1 1 ]Note that A is almost the Laplacian
but with the (n,n) th entry = 1 insteadof 2.However, I found
out that the AGM works well for the following matrix[ 2 -1 -1
2 -1 -1 2 -1.... ........ -1 2 ]but not A, does it make
sense?Thank you very much.===
===
Subject:
: Re: Ln typo
or....?Originator: baez@math-cl-n03.math.ucr.edu (John
Baez)>|Say for ln(ab) = ln(a) + ln(b) we take ln(a) as the
area from>|x = 1 to 2 and ln(b) as the area from 1 to 3, this
is the same as ln 2 +>|ln 3 = ln 6. The area from x = 1 to 6
isn't obviously the same as the>|sum of the two smaller areas.
What am I missing?>The area from 1 to 6 is obviously the sum of
the areas from 1 to 2>and from 2 to 6. So really you need to
show that the area from 1 to 3>and from 2 to 6 are the same,
or more generally, that the area from>1 to b and the area from
a to ab are the same.>There's a way to show it. Try to find it.
Do you know any techniques>of integration?This problem is so
fundamental that solving it with techniques of integration
learned in calculus class runs the risk of making itseem
harder than it is, and obscuring the geometry involved.How
about this:Take the region under the curve y = 1/x from x = 1
to x = a.If you stretch this region horizontally by the factor
b and squash it vertically by the same factor, you get another
region of the same area. And, this is the region under the
curve y = 1/x from x = b to x = ab. It follows thatln(ab) -
ln(b) = ln(a) - ln(1)and since ln(1) = 0, we getln(ab) = ln(a)
+ ln(b).===
===
Subject:
: Re: Help showing system is not Hamiltonian.
<15612c7d.0403141357.46cfa48a@posting.google.com>Craig,>> BUT
it seems to me that for all values of a, if a = c, b = d = 2a
the>> system is also Hamiltonian. Specifically if a = c = 2, b
= d = 2a = 4 we>> still have partial_x X + partial_y Y = 0 and
the system is thus>> Hamiltonian.In message
<15612c7d.0403141357.46cfa48a@posting.google.com>, Craig
>Thus, 2a = b = 2c = d > 0 is the full solution.Thus we pretty
much agree, except I take a to be the free parameter.>A quick
google search did not seem to say much about a
differential>equation being Hamiltonian but rather discussed
the Hamiltonian>operator and related equations.Yep, I tried
that as well before posting.David-- ===
===
Subject:
: Re: a digital
game|Oops! Forgot about Mr. Nobel's grudge against
mathematicians.It should perhaps be pointed out that Nobel's
reason for not> giving a prize in mathematics wasn't, contrary
to a popular> urban legend, due to a mathematician having an
affair with> his wife or girlfriend. This has been debunked
various times.> As far as I know there's no evidence of his
having a grudge> against mathematicians.Nobel claimed he was
choosing less theoretical areas for his> prizes. He seems to
have been interested in areas that could> help offset the
hazards of dynamite to society.And for the OP's benefit
economics was not one of these.The so called Nobel prize in
economics is nothingto do with Nobel.-- Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92;
mistakes his penis for a square root, 88-9Francis Wheen, _How
Mumbo-Jumbo Conquered the World_===
===
Subject:
: Multigrad
AgainDear all,I would like to ask if the multigrid method
works well for the Neumannboundary conditions, compared with
the performance of the multigrid methodon the Dirichlet
boundary conditions?Thank you very much.Alfred===
===
Subject:
:
Integration x^(-1)*exp(-(ax+c)^2+bx+d)Hi NG!I'm looking for a
solution to the following integralint x^-1 exp(-(a x +c )^2 +
b x + d) dx or int x^-1 exp(a x^2 + bx + d) dxDo you know the
solution or a way to find it? As I didn't find a solutionusing
Maple4 or the web pages of Wolframs
Research(http://functions.wolframs.com) I try this NG.Best
regards,Derk===
===
Subject:
: Re: Integration
x^(-1)*exp(-(ax+c)^2+bx+d)> Hi NG!I'm looking for a solution
to the following integralint x^-1 exp(-(a x +c )^2 + b x + d)
dx or int x^-1 exp(a x^2 + b> x + d) dxDo you know the
solution or a way to find it? As I didn't find a solution>
using Maple4 or the web pages of Wolframs Research>
(http://functions.wolframs.com) I try this NG.Best regards,>
DerkYou are never going to find an elementary antiderivative
because thereis none. The best you can get is a function using
the exponentialintegral.EI(x) = Integral{-x,
infinity}[(e^-t)/t]dt===
===
Subject:
: Re: Integration
x^(-1)*exp(-(ax+c)^2+bx+d)> Hi NG!I'm looking for a solution
to the following integralint x^-1 exp(-(a x +c )^2 + b x + d)
dx or int x^-1 exp(a x^2 + b> x + d) dxDo you know the
solution or a way to find it? As I didn't find a solution>
using Maple4 or the web pages of Wolframs Research>
(http://functions.wolframs.com) I try this NG.Best regards,>
DerkYou are never going to find an elementary antiderivative
because there> is none. The best you can get is a function
using the exponential> integral.EI(x) = Integral{-x,
infinity}[(e^-t)/t]dtI was going to write that, but then
concluded that Ei() was probablynot enough, even for integral(
exp((x-1)^2)/x ) .-- G. A. Edgar
http://www.math.ohio-state.edu/~edgar/===
===
Subject:
: Re:
Hypocritical Fortune Tellers (What are the odds?)> leave the
guy alone?? i've got him saying that God told him that this
is> his path in life in one post and then contradicting this
very statement in> the other. i just want to get an
explanation on that.maybe some introverted number theorist
will happen to the group and findsomething he says
altruistic.... and possibly lead him into a relationshipwith
the lord.... gods told me that im ok to smoke pot.... and that
i'llprobably be able to meet people im s'posed to talk to in
thosecircles....God is the God of everything in every
situation in everycircumstance.... step into the total
perspetive vortex man.:-D n'stuff===
===
Subject:
: Re: matrices>
|Let's suppose k is infinite (I imagine things will break
down> |for finite fields). Then det(A + tB) = 0 as a matrix
over k(t).> |Thus there is a nonzero vector over k(t) killed
by A + tB and we> |may assume that its entries are in k[t].
Call this vector> |v = v_0 + t v_1 + ... t^m v_m. Then (A +
tB) v = 0,> |so A v_0 = 0, A v_1 = - B v0, ...., A v_m = - B
v_{m-1} and 0 = B v_m.> |This gives one an idea to find a
counterexample. Let u, u' and u''> |be linearly independent
vectors in some vector space. Define a > |transformation on
this space by setting A u = 0, A u' = u'',> |B u = - u'' and B
u' = 0 (and A u'' = anything etc.). Then> |(A + tB)(u + t u') =
0 (with t an indeterminate). As far as I can> |see one can cook
up A and B so that they only annihlate > |multiples of u and u'
etc. But A + tB is singular.am i making a silly mistake? what's
the matter with a := 1 0> 0 0and b := 0 1> 0 0over any field?
(if there's a convention that makes the example work> then use
it.)oups! well, you're right. thanks anyway ...===
===
Subject:
:
Algebra vs. universal algebraAt
http://en.wikipedia.org/wiki/Universal_algebra is an algebra
is aset A together with a collection of operations on A. [...]
After theoperations have been specified, the nature of the
algebra can befurther limited by axiomsAt
http://en.wikipedia.org/wiki/Algebraic_structure is an
algebraicstructure consists of a set together with one or more
operations onthe set which satisfy certain axioms.So, what's
the difference between an algebra and an
algebraicstructure?===
===
Subject:
: Re: Algebra vs. universal
algebraWhile we're on the subject, years ago I took a course
in universal algebra from G.-C. Rota. He kept saying a how a
field is *not* an algebra, but I never got a satisfactory
answer why not. I think he was referring to the fact that we
cannot define the unary multiplicative inverse operator on the
entire set. But why not just omit said operator and assure the
existence of such inverses in the axioms?-- Stephen J.
Herschkorn herschko@rutcor.rutgers.edu===
===
Subject:
: Re: Algebra
vs. universal algebraNNTP-Posting-User:
[Uy7T/rmuedBn3gS6WnM8tafP070p4w9T]> While we're on the
subject, years ago I took a course in universal> algebra from
G.-C. Rota. He kept saying a how a field is *not* an> algebra,
but I never got a satisfactory answer why not. I think he> was
referring to the fact that we cannot define the unary>
multiplicative inverse operator on the entire set. But why not
just> omit said operator and assure the existence of such
inverses in the> axioms?I am not at all an expert in universal
algebra, but according to theweb page (wikipedia) cited by the
original poster, the idea inuniversal algebra seems to be to
do everything in terms of equationsand operators on the set.
The web page implies that phrases like forevery nonzero x
there exists an element y such that xy = 1 are not tobe used
in universal algebra -- if possible, you need to replace
thiswith equations involving operators (and if not possible,
you can'thave such conditions in your universal algebra).I
think this is also the distinction that the original poster
wasasking about: in an algebraic structure, you can impose
whateveraxioms you like, but in a universal algebra, your
axioms should bestated entirely in terms of equations.As such,
algebraic structures are more general, and more commonly
usedand studied. Universal algebras are more restrictive (and
hence Iwould expect them to be more powerful).-- J. H.
PalmieriDept of Mathematics, Box 354350
mailto:palmieri@math.washington.eduUniversity of Washington
http://www.math.washington.edu/~palmieri/Seattle, WA
98195-4350===
===
Subject:
: Re: Algebra vs. universal algebra> While
we're on the subject, years ago I took a course in universal>
algebra from G.-C. Rota. He kept saying a how a field is *not*
an> algebra, but I never got a satisfactory answer why not. I
think he was> referring to the fact that we cannot define the
unary multiplicative> inverse operator on the entire set. But
why not just omit said operator> and assure the existence of
such inverses in the axioms?'Cos that's not equational.In
universal algebra you have a bunch of operations posited
tosatisfy various relations (e.g. group or ring axioms).
Nowone can't do this for fields. The axiom that a nonzero
elementhas an inverse isn't of the form some equation holds
for all arguments.It's logically more complicated. This means
the category of fieldsdoesn't have various propeties that
catgeories of algebraic systems share.-- Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92;
mistakes his penis for a square root, 88-9Francis Wheen, _How
Mumbo-Jumbo Conquered the World_===
===
Subject:
: Re: Algebra vs.
universal algebra Adjunct Assistant Professor at the
University of Montana.>While we're on the subject, years ago I
took a course in universal >algebra from G.-C. Rota. He kept
saying a how a field is *not* an >algebra, but I never got a
satisfactory answer why not. I think he was >referring to the
fact that we cannot define the unary multiplicative >inverse
operator on the entire set. But why not just omit said
operator >and assure the existence of such inverses in the
axioms?Because the axioms must be identities, not just first
ordersentences. Fields are partial algebras (algebras with
operationsdefined only on subsets), but not algebras. In other
words, the kindof axioms you allow are axioms where all the
quantifications areuniversal, none of them are existential.So,
as a universal algebra, a group is an algebra of type
(2,1,0)i.e., with a binary operation, a unary operation, and a
zero-aryoperation; call them m, i, and e, respectively. The
Axioms arereally identities that the algebra satisfies, i.e.,
statements aboutequalities of certain composition of
operations:m(x,m(y,z)) = m(m(x,y),z) [associativity of
multiplication]m(x,e) = id [neutral element]m(x,i(x)) = e
[inverses]As sentences, these are all statements about for
all, with noexistence quantifiers. Or, you can state them as
commuting diagramswithout appealing to elements at all.If you
think of fields as commutative rings which also happen tohave
multiplicative inverses for nonzero elements, then you would
needto be able to write down the property of having inverses
as anidentity of ->rings<-, that is, as a set of equations of
the formw_1(x1,...,xn) = w_2(x1,....,xn)where w_1 and w_2 are
words in the language of ring theory onx1,....,xn; that is,
elements of the free commutative ring inx1,...,xn, which we
interpret as saying for all x1,...,xn, the resultof doing the
operations on the left hand side is the same as theresult of
doing the operations on the right hand side. There is nosuch
set of words that will isolate the fields and only the fields
from amongthe commutative rings, as far as I am aware.(For the
same reason, groups are not a subvariety of semigroups, thatis,
if you try to treat them as special kinds of algebras of type
(2),you will not really be isolating groups, but a larger
class ofsemigroups) --
==============================================================
========It's not denial. I'm just very selective about what I
accept as reality. --- Calvin (Calvin and
Hobbes)=======================================================
==================
===
Subject:
: Re: Algebra vs. universal algebra
Adjunct Assistant Professor at the University of Montana.>At
http://en.wikipedia.org/wiki/Universal_algebra is an algebra
is a>set A together with a collection of operations on A.
[...] After the>operations have been specified, the nature of
the algebra can be>further limited by axioms>At
http://en.wikipedia.org/wiki/Algebraic_structure is an
algebraic>structure consists of a set together with one or
more operations on>the set which satisfy certain axioms.>So,
what's the difference between an algebra and an
algebraic>structure?An algebra has an algebraic structure...In
an algebraic structure, you have a set of axioms specified
inadvance that your operations must satisfy. In an algebra,
all you haveare a bunch of operations (and then you may
specify axioms that aresatisfied by that algebra in an attempt
to study it and similarlydefined objects).--
==============================================================
========It's not denial. I'm just very selective about what I
accept as reality. --- Calvin (Calvin and
Hobbes)=======================================================
==================
===
Subject:
: Re: Algebra vs. universal algebra>
At http://en.wikipedia.org/wiki/Universal_algebra is an
algebra is a> set A together with a collection of operations
on A. [...] After the> operations have been specified, the
nature of the algebra can be> further limited by axioms> At
http://en.wikipedia.org/wiki/Algebraic_structure is an
algebraic> structure consists of a set together with one or
more operations on> the set which satisfy certain axioms.> So,
what's the difference between an algebra and an algebraic>
structure?In the fist case, axioms are specified. (They can be
more axioms of course). In the second, it's just
generalities.===
===
Subject:
: solving cubics modulo a primeI've
found another situation where I wish to solve a _fixed_
cubicmodulo many primes. Said primes will be in their ions,
both in size and in number, and I've got computational
efficiency foremost in my mind. (Another is harking back to
http://groups.google.com/groups?threadm=pan
.2002.11.28.23.05.09.37981%40yahoo.co.uk&rnum=2&prev=/groups%
3Fq%3Dauthor:carmody%2Bcubic%2Bmodular%26hl%3Den%26lr%3D%26ie%
3DUTF-8%26oe%3DUTF-8%26sa%3DN%26tab%3Dwg )The cubic in
question isf(x) = x^3+x^2-x+1There are 3 non-degenerate
cases.As -44 is the discriminant:(-11|p) = 1 : 0 or 3
solutions(-11|p) = -1 : 1 solutionAs per Bob Silverman's
recommendation(
http://groups.google.com/groups?selm=
20021129131822.29125.00000085%40mb-cb.aol.com )a simple
Cantor-Zassehaus is /ideal/ for the 1-solution case, as x^p-x
extracts precisely the linear factor in one shot. The cost of
the operation, using naive code is less than 6 modular
exponentiations of the same size. (Using L->R, the squaring
part costs 6 mulmods, the mult part is simply data shuffling
and a couple of adds and subs, due to the simple form of f(x).
The final GCD is 6 mulmods, a modinv, and some additions.)As in
theory Cardan's usual algorithm involves little more than
sqrtmod and a cubertmod, I thought that it would have a good
chance of being more computationally efficient for the task at
hand. However, I've hit a snag, and I'm really not feeling too
sharp at the moment, so have come a-grovelling.Using Ken's
link to http://www.hsu.edu/faculty/worthf/cubic.htmlas a
reference, I formed the depressed cubic x^3+Qx+R: x^3+x^2-x+1
--> x^3-4/3*x+38/27 So Q is -4/3, and R is 38/27(verification
in GP: gp > subst(x^3+x^2-x+1,x,x-1/3) x^3 - 4/3*x + 38/27Now
I tried to solve u^2 + Ru - Q^3/27 == 0 (mod p)u^2 + (38/27)u
+ (64/729) == 0 (mod p)Plugging this into the usual quadratic
formula gives:Firstly, b^2-4ac = (R^2+4*Q^3/27) == 44/27 (mod
p), so u == -19/27 +/- sqrt(11/27) (mod p)Now taking a real
example, p=47 gp > R 38/27 gp > Q -4/3 gp > polrootsmod(u^2 +
R*u - (Q^3/27), 47) []~i.e. no solutions exist for u.The
killer is that: gp > polrootsmod(x^3 - 4/3*x + 38/27,47)
[Mod(11, 47), Mod(37, 47), Mod(46, 47)]~i.e. there _do_ exist
roots for the original polynomialHave I boobed somewhere, or
is Cardano's formula really unable to solve the cubic modulo
primes?-- 1st bug in MS win2k source code found after 20
minutes: scanline.cpp2nd and 3rd bug found after 10 more
minutes: gethost.cBoth non-exploitable. (The 2nd/3rd ones
might be, depending on the CRTL)===
===
Subject:
: Re: solving cubics
modulo a prime> Have I boobed somewhere, or is Cardano's
formula really unable to > solve the cubic modulo primes?The
problem is that there is no requirement that 11/27 be a
quadraticresidue mod p, as indeed in this case it is not. The
requirement israther that u = b^3 mod p must have a solution
of the form b =x+y*sqrt(11/27) mod p. This situation is
similar to the fact that ifthe cubic over Q has three real
roots, then Cardan's formula findsthese roots via an imaginary
quadratic field, and there is basicallyno getting around
it.===
===
Subject:
: Re: solving cubics modulo a prime> Now taking a
real example, p=47 gp > R> 38/27> gp > Q> -4/3> gp >
polrootsmod(u^2 + R*u - (Q^3/27), 47)> []~i.e. no solutions
exist for u.The killer is that:> gp > polrootsmod(x^3 - 4/3*x
+ 38/27,47)> [Mod(11, 47), Mod(37, 47), Mod(46, 47)]~i.e.
there _do_ exist roots for the original polynomialHeh-heh!
This is the good old casus irreducibilis again!Just as if a
cubic has three real roots, Cardan's methodrequires the
extraction of a complex cube root, a cubicover F_p with three
roots in F_p requires the cube root to be taken in F_{p^2}.One
trick to split off some root is a variant of
theCantor-Zassenhaus trick you cite. Calculate the gcdof your
cubic with x^{(p-1)/2} - 1. If all its rootsare quadratic
residues or all are quadratic nonresidues,then bad luck,
otherwise one finds a linear or quadratic factor.To get around
the bad cases, take any integer k and dothe same with
(x-k)^{(p-1)/2} - 1. For random k thisshould work with
probability about 3/4.This is a well-known trick, I don't
recall who first usedit, but I'm sure Bob Silverman would
:-)-- Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlLacan,
Jacques, 79, 91-92; mistakes his penis for a square root,
88-9Francis Wheen, _How Mumbo-Jumbo Conquered the
World_===
===
Subject:
: Re: solving cubics modulo a primeThis is a
well-known trick, I don't recall who first usedit, but I'm
sure Bob Silverman would :-)In fact, my polynomial root finder
for theNumber Field Sieve does exactly this. Youcan find the
details in Knuth, Vol 2.You can lead a horse's ass to
knowledge, but you can't make him think.===
===
Subject:
: Re:
solving cubics modulo a prime> This is a well-known trick, I
don't recall who first used> it, but I'm sure Bob Silverman
would :-)> In fact, my polynomial root finder for the> Number
Field Sieve does exactly this. You> can find the details in
Knuth, Vol 2.To be honest, I don't find the Knuth reference a
particularly useful one for this problem; with a paedagogical
view, that is. I think it's only a good reference if you've
already been familiar with the solution, in which case there's
a reasonable chance you don't need it as a reference.I think
I've learnt more about CZ from my own scribblings and coding
today, and from Robin mentioning the x -> x-k trick, than I
have from rereading Knuth several times.However, you were spot
on about which target I should be aiming for, it's just that my
prefered route differed from your advised one. So thanks for
plugging it anyway.-- 1st bug in MS win2k source code found
after 20 minutes: scanline.cpp2nd and 3rd bug found after 10
more minutes: gethost.cBoth non-exploitable. (The 2nd/3rd ones
might be, depending on the CRTL)===
===
Subject:
: Re: solving cubics
modulo a primeNow taking a real example, p=47 gp > R> 38/27>
gp > Q> -4/3> gp > polrootsmod(u^2 + R*u - (Q^3/27), 47)>
[]~i.e. no solutions exist for u.The killer is that:> gp >
polrootsmod(x^3 - 4/3*x + 38/27,47)> [Mod(11, 47), Mod(37,
47), Mod(46, 47)]~i.e. there _do_ exist roots for the original
polynomialHeh-heh! This is the good old casus irreducibilis
again!> Just as if a cubic has three real roots, Cardan's
method> requires the extraction of a complex cube root, a
cubic> over F_p with three roots in F_p requires the cube root
> to be taken in F_{p^2}.Phew! I'm not going bonkers.I guessed
that I was creating an imaginary number, so to speak, and that
it would resolve itself by cancelling with a conjugateat the
final stage. I didn't follow it though though, as I decided
that I'd attack a quick CZ instead as something to fall back
on.Now work in F_{p^2} is often way more expensive than
working in F_p, and I think that I could do >2 CZs in the time
of one Cardan.That means Bob's prior advice of just using CZ
for all cases looksmost sage. Hats off to Bob.> One trick to
split off some root is a variant of the> Cantor-Zassenhaus
trick you cite. Calculate the gcd> of your cubic with
x^{(p-1)/2} - 1. If all its roots> are quadratic residues or
all are quadratic nonresidues,> then bad luck, otherwise one
finds a linear or quadratic factor.Yup, doing that already,
and a little more too to avoid some of the bad luck. If p==1
(mod 4) then (p-1)/2 is even, and you can lower the exponent
again. So the all-QR problem case is also resolvable some of
the time. (repeat until you run out of trailing 0s.) Assuming
you've got a few bytes to spare, keeping this power history
has zero computational overhead, obviously.> To get around the
bad cases, take any integer k and do> the same with
(x-k)^{(p-1)/2} - 1. For random k this> should work with
probability about 3/4.This is a well-known trick, I don't
recall who first used> it, but I'm sure Bob Silverman would
:-)Aha, I'd forgotten about the x-k permutation trick (I've
never actually coded it up before), but it certainly rings a
bell. Now I'm actaully coding it up, I'm sure I'll not forget
it again!-- 1st bug in MS win2k source code found after 20
minutes: scanline.cpp2nd and 3rd bug found after 10 more
minutes: gethost.cBoth non-exploitable. (The 2nd/3rd ones
might be, depending on the CRTL)===
===
Subject:
: Re: solving cubics
modulo a prime>> i.e. no solutions exist for u.>> [...]>> i.e.
there _do_ exist roots for the original polynomial>>>
Heh-heh! This is the good old casus irreducibilis again!>>
Just as if a cubic has three real roots, Cardan's method>>
requires the extraction of a complex cube root, a cubic>> over
F_p with three roots in F_p requires the cube root >> to be
taken in F_{p^2}.>Phew! I'm not going bonkers.Careful! All you
can conclude is that the prior posting _does notimply_ that
you're going bonkers. That's not the same as having itimply
that you're _not_ going bonkers. That's logic.dave
Contrariwise, continued Tweedledee, if it was so, it might be;
and if it were so, it would be; but as it isn't, it ain't.
That's logic. ===
===
Subject:
: Re: solving cubics modulo a prime>>
i.e. no solutions exist for u.>> [...]>> i.e. there _do_ exist
roots for the original polynomial>>> Heh-heh! This is the
good old casus irreducibilis again!>> Just as if a cubic has
three real roots, Cardan's method>> requires the extraction of
a complex cube root, a cubic>> over F_p with three roots in F_p
requires the cube root >> to be taken in F_{p^2}.Phew! I'm not
going bonkers.Careful! All you can conclude is that the prior
posting _does not> imply_ that you're going bonkers. That's
not the same as having it> imply that you're _not_ going
bonkers. That's logic.However, you'll notice that my latter
statement contains no terms implying deduction, inference or
anything vaguely syllogistic. What's the smiley for a tongue
stuck _really_ far out ;-)-- 1st bug in MS win2k source code
found after 20 minutes: scanline.cpp2nd and 3rd bug found
after 10 more minutes: gethost.cBoth non-exploitable. (The
2nd/3rd ones might be, depending on the CRTL)===
===
Subject:
: Re:
New
Knowledgehttp://www.personalviability.com/Articles/
mainArticle.htm$Cm3.1091@newsread3.news.pas.earthlink.net...>
How, when and why does *New Knowledge* come to
humankind?http://www.personalviability.com/Articles/
mainArticle.htmPeter Charlton===
===
Subject:
: Re: New Knowledge
charset=Windows-1252> How, when and why does *New Knowledge*
come to humankind?>
http://www.personalviability.com/Articles/mainArticle.htm>
Peter CharltonPeter,that is not searching. That's just
proselytizing for--- divine --- like enviro --- dead ends!::
for $$$$ ...1575Skr (173? / 120) per person.:: 1 day
lecture/seminar. Lunch & coffee includedYou might be
interested in the counter point byGoogle groups for (Uncle Al
Yahweh stick)hahahaha........ahahahanson===
===
Subject:
: How to
show that a polynoial is irreducible ?I have trying to show
that x^3-1 is irreducible in Z mod 7 [x]. Buti dont know where
to begin. Can someone give some hints i will behappy/ Brian
j===
===
Subject:
: Re: How to show that a polynoial is irreducible
?> I have trying to show that x^3-1 is irreducible in Z mod 7
[x]. But> i dont know where to begin. Can someone give some
hints i will be> happyIt is not irreducible. It has 1 as root,
so x-1 is a factor.-- Michael Knudsen===
===
Subject:
: Re: How to
show that a polynoial is irreducible ?> I have trying to show
that x^3-1 is irreducible in Z mod 7 [x]. But> i dont know
where to begin. Can someone give some hints i will be> happy/
Brian jAre you sure you want to post that on Usenet?Years from
now some poster here might decide that your post would makea
good addition to his webpage.You could be a part of another
variation of cyber wars.Then what would you do, eh?Sure *now*
you don't see yourself as important and *now* it hardlyseems
to matter.But where might you be in ten years Jesper
Hansen?And what do you do then if someone like Dik Winter
copies this post orothers you might make to their webpage?What
can you do then, eh?Maybe you should reconsider whether or not
Usenet is the environmentto make any posts considering that
people like Dik Winter *are* outthere.James Harris===
===
Subject:
:
Re: How to show that a polynoial is irreducible ?>I have
trying to show that x^3-1 is irreducible in Z mod 7 [x]. But>i
dont know where to beginTrying to prove something that is false
is alwaysdifficult.You can lead a horse's ass to knowledge, but
you can't make him think.===
===
Subject:
: Re: How to show that a
polynoial is irreducible ?> I have trying to show that x^3-1
is irreducible in Z mod 7 [x]. But> i dont know where to
begin. Can someone give some hints i will be> happyHint: 1 is
a root.Best regards,Jose Carlos Santos===
===
Subject:
: Re: When
Cayley transform is orthogonal ?>>>Let A be a n x n real
matrix such that det(I-A)=/=0 .>>>The Cayley transformation
of A is C(A)=(I-A)^{-1}(I+A).>>>How we can describe all
matrices A for which C(A) is orthogonal ?>>> All real
antisymmetric matrices. Note that I-C(A) = -2(I-A)^{-1}A,>>
and I+C(A) = 2(I-A)^{-1}, so that A = -(I+C(A))^{-1}(I-C(A)),
provided>> C(A) does not have -1 as an eignevalue. It now
follows from the >> orthogonality of C(A) that A is
antisymmetric. More generally, if>> C(A) has -1 as an
eigenvalue, then C(A) is of the form B diag(-I,D) B^{-1}, >>
for an orthogonal matrix B, and orthogonal matrices I and D
whose >> dimensions sum to n, and where -1 is not an
eigenvalue of D. Let>>> B^{-1} A B = (A_1 A_2)>> (A_3
A_4),>>> so that (-I O) = (I-A_1 -A_2 )^{-1} (I+A_1 A_2 )>>
( O D) (-A_3 1-A_4) ( A_3 I+A_4),>>> so that (-I+A_1 -A_2 D
) = (I+A_1 A_2 )>> ( A_3 D-A_4 D) ( A_3 I+A_4).>>> This
can't hold unless C(A) does not have -1 as an eigenvalue. In
other>> words, C(A) is orthogonal iff A is antisymmetric, and
in that case, >> C(A) does not have -1 as an eigenvalue.>>>
David McAnally>>> But I'm always true to you, darlin', in my
fashion,>> Yes, I'm always true to you, darlin', in my way.>>
-- Lois Lane>>> ---->Thank you for your nice solution,
AlexYou're welcome :-)David McAnally But I'm always true to
you, darlin', in my fashion, Yes, I'm always true to you,
darlin', in my way. -- Lois Lane----===
===
Subject:
:
testtest===
===
Subject:
: Test for the 15/03/04This a
test===
===
Subject:
: Analysis questionI encounter following 2
difficult(to me) problems.Can anybody help me?The first
problem is :If f:R->R satisfies f(x+y)=f(x)+f(y) for all x,y
in R and lim {x->0} f(x) = L ,Then prove L = 0. I know If f is
continuous on R then f(x)=f(1)x for all x in R.So, If f(x) is
continous on R then I can prove this.But, this problem does
not have continuity condition andso, I don't know how can I
prove it.Question 2>If f:R->R satisfies f(x+y)=f(x)+f(y) for
all x,y in R and continuous at x=c in R, then How can I prove
f(x) is continous on every point x in R?If somebody can help
me, Please post reply.===
===
Subject:
: Re: Analysis questionI
encounter following 2 difficult(to me) problems.> Can anybody
help me?> The first problem is :> If f:R->R satisfies
f(x+y)=f(x)+f(y) for all x,y in R and lim {x->0} f(x)> = L ,
Then prove L = 0.I know If f is continuous on R then
f(x)=f(1)x for all x in R.> So, If f(x) is continous on R then
I can prove this.> But, this problem does not have continuity
condition and> so, I don't know how can I prove it.Question 2
If f:R->R satisfies f(x+y)=f(x)+f(y) for all x,y in R and
continuous at> x=c in R, then How can I prove f(x) is
continous on every point x in R?Perhaps this will
help:consider R as a vector space over Q. Then f is an
endomorphism.The following is well-known for vector spaces
over R:- a linear map is continuous in 0 iff it is continuous
everywhere iff f mapsbounded sets to bounded sets.If you look
at the proof, you will see that it holds equally well for
vectorspaces over Q.-- I'm on warm milk and
laxativesCherry-flavored antacidsreverse my forename for mail!
- saibot===
===
Subject:
: Re: Analysis question> The first problem is
:> If f:R->R satisfies f(x+y)=f(x)+f(y) for all x,y in R and
lim {x->0} f(x) = L ,> Then prove L = 0. 2L = lim_{x->0} 2f(x)
= lim_{x->0} f(2x) = lim_{x->0} f(x) = L.2L = L => L=0.> If
f:R->R satisfies f(x+y)=f(x)+f(y) for all x,y in R and
continuous at x=c in R, > then How can I prove f(x) is
continous on every point x in R?lim_{x->0} f(x) = lim_{x->0}
f((x + c) - c) = = lim_{x->0} f(x + c) + lim_{x->0} f(-c) =
f(c) + f(-c) = f(0) = 0.So, f is continuous at 0.
Now,lim_{x->a} f(x) = lim_{x->a} f(x - a) + lim_{x->a} f(a) =
= f(0) + f(a) = f(a).Best regards,Jose Carlos
Santos===
===
Subject:
: Re: Analysis question> lim_{x->a} f(x) =
lim_{x->a} f(x - a) + lim_{x->a} f(a) => = f(0) + f(a)> =
f(a).Doesn't saying that lim x->a f(x-a) = f(0) imply
continuity in the firstplace?===
===
Subject:
: Formula for Pi by
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id i2FBLp832461;Formula for PiSeveral
years ago, while at high school, I derived the following
formula for Pi:Pi = lim 2^(n+1)*sqrt(2 - sqrt(2 + sqrt(2 +
sqrt(2 + ...)))). n->infinity <----------- n times
----------->The beauty of this formula, I think, is that it
can be derived using only Pythagoras formula.Now I wonder if
it already exists in the mathematical literature. Probably it
does. If affirmative, I would appreciate any somebody telling
me where.John Fredsted===
===
Subject:
: Re: Formula for Pi> Formula
for PiSeveral years ago, while at high school, I derived the
following formula for Pi:Pi = lim 2^(n+1)*sqrt(2 - sqrt(2 +
sqrt(2 + sqrt(2 + ...)))).> n->infinity <----------- n times
----------->The beauty of this formula, I think, is that it
can be derived using only Pythagoras formula.Now I wonder if
it already exists in the mathematical literature. Probably it
does. If> affirmative, I would appreciate any somebody telling
me where.Your formula is formula (44)
fromhttp://mathworld.wolfram.com/PiFormulas.htmlBest
regards,Jose Carlos Santos===
===
Subject:
: Re: Please help with
Haar Integration>> Two suggestions:>> 1) Try to visualize how
does sum (from j = 1 to m) c_ j 1_n(x_ j + x)>> look like.>>
2) Don't forget that f has compact support.I still do not see
this at all. This means for any real-valued function> with
compact support, we can really bound f(x) like this?No. But
the answer is affirmative if you add continuous.> I just can't
see> this. But for a moment, forget that I can't see this and
note that I can> understand (I think) that (f_0 : 1_n) = n ,
which I talk about below. If I> understand this, then can I
just say that since we can just bound f(x) by> step functions,
the above result is true? (Read below) > Now, let (f : 1_n) be
the following infimum : inf{sum (from j = 1 to n)> c_ j>> such
that c_1,...,c_n > 0 and there exist x_1,...,x_n in R such
that> f(x)>> <= sum (from j = 1 to n) c_ j 1_n(x_ j + x)}.That
last sentence I copied directly from the book and you now can
see> that>> the author's indices are mixed up here.>> You're
right here. What book is this?>>The book is A first course in
Harmonic Analysis by Anton Deitmar. It is> trying to define
Haar integral by motivating a different definition of> Riemann
integral. Could you please tell me which indices are mixed up
so I> can understand? Also please read further I explain a lot
more.I think that the definition should beinf{sum (from j = 1
to m) c_ j such that c_1,...,c_m > 0 and there exist
x_1,...,x_m in R such that f(x) <= sum (from j = 1 to m) c_ j
1_n(x_ j + x)}.So, every _n_ is replaced by _m_, except in
1_n. I have alreadywritten to professor Deitmar to ask me
whether or not I am right here,half an ahour ago).>> Then he
says that the Riemann integral can be described asIntegral
(from - infinity to infinity) f(x) dx = lim (n ---> infinity)>
of>> (f : 1_n)/nWhy is this true?>> I do not understand the
meaning of (f : 1_n). Could you please explain?the meaning of
(f : 1_n).> I only copied it from my book...I don't think I
understand it either.Oh, but I do! :-)Best regards,Jose Carlos
Santos===
===
Subject:
: Re: Please help with Haar Integration>> Two
suggestions:1) Try to visualize how does sum (from j = 1 to m)
c_ j 1_n(x_ j + x)>> look like.2) Don't forget that f has
compact support. I still do not see this at all. This means
for any real-valued function> with compact support, we can
really bound f(x) like this?> No. But the answer is
affirmative if you add continuous.> I just can't see> this.
But for a moment, forget that I can't see this and note that
Ican> understand (I think) that (f_0 : 1_n) = n , which I talk
about below.If I> understand this, then can I just say that
since we can just bound f(x)by> step functions, the above
result is true? (Read below)> Now, let (f : 1_n) be the
following infimum : inf{sum (from j = 1 ton)> c_ j>> such that
c_1,...,c_n > 0 and there exist x_1,...,x_n in R such that>
f(x)>> <= sum (from j = 1 to n) c_ j 1_n(x_ j + x)}.That last
sentence I copied directly from the book and you now cansee>
that>> the author's indices are mixed up here.You're right
here. What book is this? The book is A first course in
Harmonic Analysis by Anton Deitmar. Itis> trying to define
Haar integral by motivating a different definition of> Riemann
integral. Could you please tell me which indices are mixed upso
I> can understand? Also please read further I explain a lot
more.> I think that the definition should be> inf{sum (from j
= 1 to m) c_ j such that c_1,...,c_m > 0 and there> exist
x_1,...,x_m in R such that f(x) <= sum (from j = 1 to m) c_ j>
1_n(x_ j + x)}.> So, every _n_ is replaced by _m_, except in
1_n. I have already> written to professor Deitmar to ask me
whether or not I am right here,> half an ahour ago).>> Then he
says that the Riemann integral can be described asIntegral
(from - infinity to infinity) f(x) dx = lim (n --->infinity)>
of>> (f : 1_n)/nWhy is this true?I do not understand the
meaning of (f : 1_n). Could you please explain?> the meaning
of (f : 1_n).> I only copied it from my book...I don't think I
understand it either.> Oh, but I do! :-)Please reply with your
findings from Prof. Deitmar and with an explanationof what the
heck is goingthrough the whole thing, even the firstpart about
how we can bound f(x) that is continuous with compact
support(compact support here means thatf(x) >= 0 for all x in
R, correct?)> Best regards,> Jose Carlos Santos===
===
Subject:
: Re:
A Paradox of The Central Limit Theorem> Sorry, I meant
lim_{n->infinity}Pr(f(X_n)) not lim_{n->infinity}f(n).Really?
To quote you again Well, in my view, the key point of this
paradox is that> lim{n->infinity}Pr{f(n)} =
Pr{lim{n->infinity}f(n)} does > not necessarily hold. How do
you say about this?So you meanlim{n->infinity}Pr{f(n)} =
lim_{n->infinity}Pr(f(X_n))> Non. I mean
lim_{n->infinity}Pr(f(X_n)) = Pr(lim_{n->infinity}f(X_n))does
not necessarily hold true ---- It is a matter of fact
though.===
===
Subject:
: Re: A Paradox of The Central Limit
Theorem>>>> Sorry, I meant lim_{n->infinity}Pr(f(X_n))
not>> lim_{n->infinity}f(n).>>> Really? To quote you again>> Well, in my view, the key point of this paradox is that>>
lim{n->infinity}Pr{f(n)} = Pr{lim{n->infinity}f(n)} does>> not
necessarily hold. How do you say about this?>>> So you mean>> lim{n->infinity}Pr{f(n)} = lim_{n->infinity}Pr(f(X_n))>Non.I
mean lim_{n->infinity}Pr(f(X_n)) = Pr(lim_{n->infinity}f(X_n))>
does not necessarily hold true ---- It is a matter of fact
though.I supposed that f denoted some kind of event. If that
is so,then lim_{n->infinity}f(X_n) is a limit of events and
mustbe an event itself. I didn't know there was a theory of
limitsof events --- so what do you mean by
lim_{n->infinity}f(X_n)?lim_{n->infinity}f(X_n)-- Robin
Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79,
91-92; mistakes his penis for a square root, 88-9Francis
Wheen, _How Mumbo-Jumbo Conquered the World_===
===
Subject:
: q:
linear algebra / matricesHelloJust taking the Linear algebra
course and got a question:Is it possible to find the inverse
of a matrix A given only: adj(A) ?Perhaps one can reconstruct
A (from adj(A)) to compute det(A) andcomplete the
following:A^-1 = 1/det(A) * adj(A)Many matrices can have the
same adj() ?If adj(A)==adj(B) does this means det(A) == det(B)
?--Elias===
===
Subject:
: Re: q: linear algebra / matrices>Hello>Just
taking the Linear algebra course and got a question:>Is it
possible to find the inverse of a matrix A given only: adj(A)
?You might like to consider A = -I in odd dimensions.Not sure
about even dimensions though.Derek Holt.>Perhaps one can
reconstruct A (from adj(A)) to compute det(A) and>complete the
following:>A^-1 = 1/det(A) * adj(A)>Many matrices can have the
same adj() ?>If adj(A)==adj(B) does this means det(A) ==
det(B) ?>-->Elias===
===
Subject:
: Re: q: linear algebra /
matrices>>Hello>>Just taking the Linear algebra course and got
a question:>>Is it possible to find the inverse of a matrix A
given only: adj(A) ?>You might like to consider A = -I in odd
dimensions.>Not sure about even dimensions though.I think the
general answer to your question for nxn matrices over a fieldK
is yes if and only if K contains no (n-1)-st roots of 1 other
than 1.Note that if w is an (n-1)-st root of 1, then adj(wI) =
adj(I) = I.On the other hand, det(A)^(n-1) = det(adj(A)), so
det(A) is determinedby adj(A) up to multiplication by an
(n-1)-st root of 1.So over the real field, it is yes for all
even n, but over the complexfield it is no for all n >
2.>>Perhaps one can reconstruct A (from adj(A)) to compute
det(A) and>>complete the following:>>A^-1 = 1/det(A) *
adj(A)>>Many matrices can have the same adj() ?Note that any
matrix of rank < n-1 has adj(A) = 0.Derek Holt.>>If
adj(A)==adj(B) does this means det(A) == det(B)
?>>-->>Elias===
===
Subject:
: Re: q: linear algebra /
matrices|>>Hello|>>|>>Just taking the Linear algebra course
and got a question:|>>|>>Is it possible to find the inverse of
a matrix A given only: adj(A) ?|>|>You might like to consider A
= -I in odd dimensions.|>Not sure about even dimensions
though.||I think the general answer to your question for nxn
matrices over a field|K is yes if and only if K contains no
(n-1)-st roots of 1 other than 1.||Note that if w is an
(n-1)-st root of 1, then adj(wI) = adj(I) = I.|On the other
hand, det(A)^(n-1) = det(adj(A)), so det(A) is determined|by
adj(A) up to multiplication by an (n-1)-st root of 1.i guess
i'm hoping that the only reason i can't figure out
whatquestion it is that you're discussing here is because i
have no ideawhat adj stands for here.-- [e-mail address
jdolan@math.ucr.edu]===
===
Subject:
: Inner productsDear math
experts, please help with the following question :My text says
this :Let K be a compact metrizable group and f be a
representation on afinite-dimensional hilbert space (V,
<*,*>). Suppose that we can show thatthere is a second inner
product (*,*) on V such that K is unitary withrespect to
(*,*). Since every inner product on V is of the form (v,w)
=<Sv,Sw> for some S in GL(V), .....Why is the last part true?
Every inner product on V is of the form (v,w) =<Sv,Sw> for
some S in GL(V)? I understand that (v,w) = <Sv,Sw> is an
innerproduct, that makes sense, but every inner product is of
that form?Moshe===
===
Subject:
: Re: Inner products>My text says
this :>Let K be a compact metrizable group and f be a
representation on a>finite-dimensional hilbert space (V,
<*,*>). Suppose that we can show that>there is a second inner
product (*,*) on V such that K is unitary with>respect to
(*,*). Since every inner product on V is of the form (v,w)
=><Sv,Sw> for some S in GL(V), .....>Why is the last part
true? Every inner product on V is of the form (v,w) =><Sv,Sw>
for some S in GL(V)? I understand that (v,w) = <Sv,Sw> is an
inner>product, that makes sense, but every inner product is of
that form?Let n be the dimension of V. Let {u_j:1<=j<=n} be an
orthonormal basisof (V,(*,*)) and {v_j:1<=j<=n} an orthonormal
basis of (V,<*,*>). Sincea basis is linearly independent, there
is a matrix S so that Su_j = v_j.Since (u_i,u_j) = I =
<v_i,v_j> = <Su_i,Su_j> and (*,*) is bilinear, itis easy to
see that (v,w) = <Sv,Sw> for all v,w in V.Rob Johnson
<rob@trash.whim.org>take out the trash before
replying===
===
Subject:
: Re: Inner products> Dear math experts,
please help with the following question :My text says this
:Let K be a compact metrizable group and f be a representation
on a> finite-dimensional hilbert space (V, <*,*>). Suppose that
we can show> that there is a second inner product (*,*) on V
such that K is unitary> with> respect to (*,*). Since every
inner product on V is of the form (v,w) => <Sv,Sw> for some S
in GL(V), .....Why is the last part true? Every inner product
on V is of the form (v,w)> => <Sv,Sw> for some S in GL(V)? I
understand that (v,w) = <Sv,Sw> is an> inner product, that
makes sense, but every inner product is of that form?Certainly
each sesquilinear form can be expressed as(v,w) |--> <v,Aw>for
some matrix A. This form is unitary iff A is Hermitian and
positivedefinite.Now <Sv,Sw> = <v S^H S w> where S^H is the
Hermitian conjugate of S.Now to show that <v,Aw> = <Sv,Sw> for
all v and w, you gottashow that each positive definite A has
the form S^H S for somenonsingular S.-- Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92;
mistakes his penis for a square root, 88-9Francis Wheen, _How
Mumbo-Jumbo Conquered the World_===
===
Subject:
: Least square soln
of affine transform, t is given by centroid difference ?It is
intuitive to say that, for 2D affine transforms, X' = AX +
twhere X' and X are both N by 2 vectors, N being the number of
2points,the best estimate of t is given by the difference
vector between thecentroid of the two point sets X and X'.But
I cannot see this from the least square solution, although I
thinkit can be proven.===
===
Subject:
: Re: Plotting f(x,y)=0> Surf
worked well when I last used it and it's even open source: >
<http://surf.sourceforge.net>Oops, I'm a Minion of Evil [tm]
:-) (But it was the decisionof my prof to go Windows :-)--
Hauke Reddmann <:-EX8 fc3a501@uni-hamburg.deals man ankam
wollte man werden, die geschichte schreiben,die doofen sollen
sterben, der plan als man damals nach hamburg
kam(Kettcar)===
===
Subject:
: Re: Royden qn by
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id i2FEFnM21454;oh for 2) I was able
to prove that my space has no convex open sets other than R &
the empty set so I think that gives me 2) by def of locally
convex space.I'm still stumped as to even where to begin the
other qn, qn 1). Any aid would be appreciated===
===
Subject:
:
re:Antidiagonal, InfinityYeah, Ross, when you finish building
a bijection between the reals andthe integers, spare some time
to find a method of squaring the circleand if it's not too
difficult, an algorithm of solving generaldiophantine
equations. ===
===
Subject:
: Re: re:Antidiagonal, Infinity> Yeah,
Ross, when you finish building a bijection between the reals
and> the integers, spare some time to find a method of
squaring the circle> and if it's not too difficult, an
algorithm of solving general> diophantine equations.... but
first let him prove that there is *no* bijectionbetween N and
Z ;-)===
===
Subject:
: Re: Antidiagonal, Infinity> Yeah, Ross, when
you finish building a bijection between the reals and> the
integers, spare some time to find a method of squaring the
circle> and if it's not too difficult, an algorithm of solving
general> diophantine equations.... but first let him prove that
there is *no* bijection> between N and Z ;-)Dirk, you're a
troll.{1, 10, 11, ...} in binary is a completely different set
than the onein decimal.You should well know that in discussion
of decimal that those twodifferent sets have what I call PC
for Post Cantoriancardinalities that differ. That is just to
reuse the word,cardinality, I am content to not and say that
one setminus the otheris non-empty but not vice-versa or that
in any sufficiently largeinterval of their lowest ceiling set
the naturals that there is moreof one than the other or that
one set has infinitely many propersubsets that are proper
supersets of the other. I also am quite clearin that those two
sets, or indeed the one set in two differentnotations have
exactly the same cardinal number, and those twodifferent sets
have different relaxed or nonstandard asymptoticdensities.Why
don't you go back to quietly playing with your marbles.
Perhapsyou can derive a combinatoric explanation of the
Borelmeasure/asymptotic combinatorial discrepancy of numbers
in binary withvarious ratios of ones and zeros in any
sufficiently large contiguousfinite sequence in their
description. Do you remember the analogy ofthe marble
toy?http://groups.google.com/groups?selm=3B3B7FB1.E07F9CD8%
40calpha.comCalling someone incompetent isn't necessarily an
insult. There arevarious measures of incompetence, for example
the idiot scale, Dirk.Do you have anything constructive to
add?Your non-sequitur: The leap is many thousands of muscle
cells actingin unison, kilos of Avogadros. (Avogadro's number
is dimensionless.) Sometimes you can leap higher if you land
harder. That's aboutgravity. Compress a spring and then
release it, it will then leapthrough the air. Gravity is
unchanged.It is Galileo to whom is accredited that functions
exist betweeninfinite sets. It's also well-known and
trivial.The diophantine solutions are already known. To
trisect an angle withrule and compass only takes infinitely
many bisections. Perhaps tosquare the circle it takes only
infinitely many.Ross F.===
===
Subject:
: Re: Antidiagonal, Infinity>
{1, 10, 11, ...} in binary is a completely different set than
the one> in decimal.Than which one?You should well know that
in discussion of decimal that those two> different sets have
what I call PC for Post Cantorian> cardinalities that
differ.There is no way that any countably infinite set can
differ in cardinality from any other countably infinite set,
whether pre or post.> Why don't you go back to quietly playing
with your marbles. At least those whom you are criticizing have
not lost theirs. [ Irrelevancy Deleted ] Calling someone
incompetent isn't necessarily an insult. There are> various
measures of incompetence, for example the idiot scale,
Dirk.But Ross sets new records for incompetence and idiocy
almost every time he posts.===
===
Subject:
: Re: Antidiagonal,
Infinity <9Oj5c.852$Fl.427@news.cpqcorp.net> Discussion,
linux)>> Yeah, Ross, when you finish building a bijection
between the reals and>> the integers, spare some time to find
a method of squaring the circle>> and if it's not too
difficult, an algorithm of solving general>> diophantine
equations.> ... but first let him prove that there is *no*
bijection> between N and Z ;-)I think that's 's thing.But I
often get my Rosses mixed up with my s, so don't take myword
for it.-- That's all the legacy I ever wanted, to have people
remember me likea shooting star streaking across their Life
sky, illuminating, forjust one moment, unparalleled beauty
unique to itself. -- Weblogs are a particularly humble medium,
unique to themselves.===
===
Subject:
: Re: Antidiagonal, Infinity>>
Yeah, Ross, when you finish building a bijection between the
reals and>> the integers, spare some time to find a method of
squaring the circle>> and if it's not too difficult, an
algorithm of solving general>> diophantine equations. ... but
first let him prove that there is *no* bijection> between N
and Z ;-)> I think that's 's thing.> But I often get my Rosses
mixed up with my s, so don't take my> word for it.Let's take
Ross' word for it:
http://groups.google.com/groups?&threadm=blRr6.195$tk.45219@
afrodite.telenet-ops.be;-)===
===
Subject:
: Re: Antidiagonal,
Infinity <9Oj5c.852$Fl.427@news.cpqcorp.net>
<87oeqxkexe.fsf@phiwumbda.org>
<sYo5c.36388$kk2.2308156@phobos.telenet-ops.be> Discussion,
linux)> Yeah, Ross, when you finish building a bijection
between the reals and> the integers, spare some time to find a
method of squaring the circle> and if it's not too difficult,
an algorithm of solving general> diophantine equations.... but
first let him prove that there is *no* bijection>> between N
and Z ;-)>> I think that's 's thing.>> But I often get my
Rosses mixed up with my s, so don't take my>> word for it.>
Let's take Ross' word for it:>
http://groups.google.com/groups?&threadm=blRr6.195$tk.45219@
afrodite.telenet-ops.beOh, like *he* knows. Please.-- Well, I
don't claim to be an expert, in fact I am a fry cook with
anational burger chain, but I have solved many differential
and partialdifferential equations numerically. --C.
Bond===
===
Subject:
: Re: Antidiagonal, Infinity> I have a strong
interest in mapping between > the naturals and the unit
interval of the reals without discarding> ZF, > in attempts to
unite (unify, unitize) the analog and digital.> Good luck.Luck
isn't the half of it, if I am to rationally go about
workingwithin a set-theoretic foundational framework to
discuss measuredlymapping the integers to and from the unit
interval of reals.I want to understand why the infinitesimal
calculus is able tocorrespond to the consideration of the
reals as points and what itmeans for a line made only of reals
as points is able to have theproperty of continuity.One
conclusion is that the reals within the interval are
one-sided,this has to do with when I pointedly declare
statements about aperceived analogy between the reals and
rationals and samplingfrequency and frequency.Consider the
function f(x)=1, defined on the reals, it's
univariatenon-complex integral S 1 dx is x. The definite
integral a^S^b dxevaluates to b-a, on [0,1] 1-0=1.When I
integrated f(x)=1 defined on all of the naturals, I got
aresult equal to two instead of the result that I expected,
one. Now,obviously enough the integral or infinitesimal
calculus is not summingthe area from each point of a curve
above the axis, except that itactually is. We have the very
functional concept of the limit whichallows the derivation of
the integration formulae with the f(x)-f(x+1)to get the
correct symbolic results of for example the indefiniteintegral
of f(x)=1 dx to be equal to x. See Points on a
Line,Infinity.There might never be two points that are
continuous, but only at leastthree.There are lots of
variables. I'm not a polymath, I don't have the experience in
a very wide rangeof mathematical disciplines to form perhaps
the most lucid andhistorically defined explanation of the
continuous reals as points ona line, what I have are some
simple tools of integral calculus andsome depth of
consideration of points on a line and the
set-theoreticfoundations behind them, separately from various
considerations ofaxiom-free theories of mathematical
foundations. I think f(x)=1defined on the naturals is equal to
two, and that f(x)=1 defined onthe reals over [0,1] is equal to
one, and the reals are only points.With these simple tools, and
various semi-sophisticated perceptions, Icome across something
like the impulse function which is designed ordescribed to
have a value of infinity of zero and zero elsewhere onthe
reals an that's its integral evaluates to one, it is not
exactlythat. I consider the coordinate system a square matrix
and stack thef(x)=1 points to form the value at zero, it
should have the samearea as for example how a square halved
vertically with the halvesstacked has the same area. It does
not, unless the impulse functiondefined on f(x)=[0,oo) and
[0,-oo) each integrally evaluate to one,summing to two. I
focussed on the impulse function becuase it isamong a very
small class of functions that I have seen described thatis
zero at all points except one where it is defined to be
infiniteand its indefinite integrals are known. Someone here
should explainmuch about the unit impulse function.So when I
considered the impulse function and its integration equal
toone, instead of two, I got to thinking that one of its
dimensions hadbeen halved, and since it was point width, its
extent was only halfinfinity. Another way to consider it is
that it is only defined on[0,oo), so only one side of the
point gets considered.When I first posted to sci.math, I
thought I had some very deepthoughts about infinity. Then, I
heard about Cantor, and set theory,and after cursory
examination, claimed that infinite sets wereequivalent. Over
the years, I discovered some things about settheory, infinite
sets are equivalent by fiat where necessary orconvenient to
quell disclaim, and the deep superquantum is the
scalarinfinity and the subquantum the scalar
infinitesimal.Anyways about the antidiagonal argument,
Cantor's results aboutmapping sets are strong ones, and to
extend them compositionally is apleasure, obsolescence.Warm
regards, Ross F.===
===
Subject:
: Re: Antidiagonal, Infinity> Luck
isn't the half of it, if I am to rationally go about working>
within a set-theoretic foundational framework to discuss
measuredly> mapping the integers to and from the unit interval
of reals.Luck is all of it, as far as Ross' attempts to connect
with the reals is concerned. And so far all bad.===
===
Subject:
: Re:
Antidiagonal, Infinity>I think f(x)=1 defined on the naturals
is equal to twoNo thought is in evidence.-- I'm not interested
in mathematics that might have anythingto do with reality. --
Easterly, in sci.math===
===
Subject:
: Irreducible polynomials
yielding isomorphiq field extensionsThe following question was
asked some months ago in de.sci.mathematik and remained
unanswered there:Let K be a field and p, q be irreducible
monic polynomials over K. Are there simple criteria for p, q
giving isomorphic field extensions?Boudewijn===
===
Subject:
: Re: So
WTF, I think you're the Truman too>>No, no - he actually has
something approaching workable>>on this one - they do it with
Lasers. Of course he didn't bother>>to explain how they manage
to power lasers big enough to>>punch through air, clouds & such
and still be able to produce>>audio effects. I don't think that
solar power is up to that task.>>Especially if he's claiming
that they are providing this stuff to>>100,000 folk - it might
well scan that fast, but the power drain>>goes up accordingly.>
too many losers responding, I'll pick one at random, I need a
new button> on Outlook[New Post] [Reply Group] [Reply Sender]
[Reply Loser]You think we're losers? We're not the ones
imagining that everyone is listening to our unemployed weight
lifting antics via laser beams from space.> Its a laser, the
weather satellites scan 100s of kilometers long lines> through
clouds. There would be some limit to the area for generating>
sound but a town is only 10km by 10km, and it isn't 24 hours
broadcast for the> whole town, its probably distributed at
different areas at different times.By definition a laser would
bounce off the top of a building. And if someone produced a
radio beam that could pass straight through the building, it
would pass through you as well. And unless you and the
building had a reflective mirror exactly perpendicular to the
laser the beam would scatter off in random directions and not
back up to the receiver on the satellite. And unless you
happened to be shouting up at the roof space, it couldn't even
pick up your voice (assuming you were talking). And if you were
in a busy area, or in a shopping mall, or multi-storey
building, it would pick so much ambient noise to produce a
cacophony.In other words, your amazing laser theory is junk.
It wouldn't work, it is impossible, it is fantastic, it is
ludicrous.> But everyone in the town is fully clued up on
everything about me, so they're> getting a reasonable
broadcast, certainly when I'm near.HercI have to wonder why
you're even talking when there is no one within earshot to
hear it. If you are getting peculiar looks from other people,
I suggest a more straightforward reason is that anyone
speaking to themselves is a weirdo.Are you going to see a
psychiatrist?===
===
Subject:
: Re: So WTF, I think you're the Truman
too> I don't want to burst his widdle bubble, but sound waves
require a medium,> and a vacuum precludes any sort of sound. I
bet he thinks that the Shuttle> has a deep bass rumble while in
orbit.But it does in the films!===
===
Subject:
: Re: LatLong inside
polygon <7HgQpIAtkKVAFw0X@lampsacos.demon.co.uk>[probably
better than my solution because mine requires you to check
thesign of a vector]>As a quick check, choose a kind of
mid-point of the four>corners, as follows:> P is the midpoint
of diagonal AC> Q is the midpoint of diagonal BD> M (our
midpoint) is the mid-point of PQ>Had ABCD been in a plane, we
would have P=Q=M, and if>ABCD is not __too__ big compared with
the radius of>the Earth, or if they are __nearly__ on a small
>circle, they won't be that far apart.ABCD will always be in a
plane. This is true not just for squares but anyregular polygon
drawn on the surface of a sphere. Proof: The first 3 vertices
are lie in a plane (as do any 3 points).Consider the next
vertex. If it were not in the plane, the figure would
beasymetric w.r.t. 360/n degree rotation about the center, and
therefore not aregular polygon. Therefore that vertex is in the
same plane.Repeat until you have considered all
vertices.--Keith Lewis klewis {at} mitre.orgThe above may not
(yet) represent the opinions of my employer.===
===
Subject:
: Re:
LatLong inside polygon
<7HgQpIAtkKVAFw0X@lampsacos.demon.co.uk>
<c34hej$sf8$1@newslocal.mitre.orgABCD will always be in a
plane. This is true not just for squares but any>regular
polygon drawn on the surface of a sphere. Please define what
you mean by 'regular polygon' in this (3-D) context.The
original definition of a 'regular polygon' comes from
Planegeometry, in which case your result is too obvious to
beinteresting.If you insist on equal lengths of sides and
equal angles at every vertex, your theorem is not true. It
might be a 6-gon, zig-zaggingaround the equator with
rotational symmetry of order 3, for example. Or it might be a
8-gon with rotation symmetry of order two,and all edges
meeting at right-angles.If you insist on an n-gon with
rotational symmetry of order n, weare indeed back at the PLANE
definition.---------------As regards the OP's question, I had
assumed that he put square in quotes, as square because he
didn't really mean it. - -Ken, __O -<,_ (_)/ (_)Virtuale
Saluton.===
===
Subject:
: Re: Is this travelling salesman problem
version known?> If you are required to return to the starting
point,> then this is the Hamiltonian cycle problem,> which is
knownto be NP-hard.No. The path is allowed to be open (and in
the problem I'm working on,> it is always open except we have
only one city)I din't even see a proof of NP-hardness for the
open path version of> TSP. Here help is appreciated
too.Hamiltonian Path. NP-complete - easy trafo from HC or TSP.
Cheers Tuomo===
===
Subject:
: Re: Packing squares into a 1 x Zeta[2]
rectangle> Does anyone know about the more general problem of
packing n-cubes of> side-lengths 1, 1/2, 1/3,... into an n-box
of n-volume zeta(n)?You can't fit a cube of side length 1 and a
cube of sidelength 1/2 inside a right-angled box of volume less
than 3/2.--J K Hauglandhttp://www.neutreeko.com===
===
Subject:
: Re:
Packing squares into a 1 x Zeta[2] rectangleDoes anyone know
about the more general problem of packing n-cubes of>
side-lengths 1, 1/2, 1/3,... into an n-box of n-volume
zeta(n)?You can't fit a cube of side length 1 and a cube of
side> length 1/2 inside a right-angled box of volume less than
3/2.DOH!I like David W Cantrell's suggestion in the other reply
to JanKristian Haugland's post.Leroy Quet===
===
Subject:
: Re:
Packing squares into a 1 x Zeta[2] rectangle> Does anyone know
about the more general problem of packing n-cubes of>
side-lengths 1, 1/2, 1/3,... into an n-box of n-volume
zeta(n)?> You can't fit a cube of side length 1 and a cube of
side> length 1/2 inside a right-angled box of volume less than
3/2.Exactly, noting that zeta(n) < 3/2 for n > 2.However,
interesting questions might still be obtained for dimensionsn
> 2 by omitting the first few largest cubes. As an example,
let'sconsider n = 3:Omitting just the cube of edge length 1
can't work [because zeta(3) - 1 isstill smaller than
(1/2)^2*(1/2 + 1/3) ]. But omitting the cubes of edgelengths 1
and 1/2 might give an interesting packing problem
[becausezeta(3) - 9/8 is now larger than (1/3)^2*(1/3 + 1/4)
]. So I would guessthat cubes of edge lengths 1/3, 1/4, 1/5,
... might be packed into a box ofedge lengths 1/3, 1/3,
zeta(3) - 9/8. Furthermore, if such a packing ispossible, I
would suppose that a very simple algorithm could be used
todescribe how such a packing could be achieved.David
Cantrell===
===
Subject:
: Re: Packing squares into a 1 x Zeta[2]
rectangle> Since the sum of the reciprocals of the squares of
the positive integers> is pi^2/6, the question arises as to
whether squares with sides 1, 1/2,> 1/3, etc can be packed
into a rectangle of size 1 by pi^2/6....> uses exact
arithmetic throughout. The algorithm is simple:But as a matter
of interest,how did you deal with pi^2 using exact
arithmetic?-- Timothy Murphy e-mail (<80k only): tim /at/
birdsnest.maths.tcd.ietel: +353-86-2336090,
+353-1-2842366s-mail: School of Mathematics, Trinity College,
Dublin 2, Ireland===
===
Subject:
: Re: Packing squares into a 1 x
Zeta[2] rectangle> Since the sum of the reciprocals of the
squares of the positive integers> is pi^2/6, the question
arises as to whether squares with sides 1, 1/2,> 1/3, etc can
be packed into a rectangle of size 1 by pi^2/6.> ...> uses
exact arithmetic throughout. The algorithm is simple:> But as
a matter of interest,> how did you deal with pi^2 using exact
arithmetic?Mathematica does it all for me. Mathematica can do
comparisons between, forexample, rational numbers and
transcendentals. Here is a short Mathematicasession in which
it compares a rational, num/denom, with pi+e and prints
theresult Greater, it then compares the rational (num-1)/denom
with pi+e andprints the result Not greater:In[1]:=
$MaxExtraPrecision = 100Out[1]= 100In[2]:= num =
6083355812763767537862691282377375125710417240146Out[2]=
6083355812763767537862691282377375125710417240146In[3]:= denom
= 1038137562741240040724843633763265660486107102341Out[3]=
1038137562741240040724843633763265660486107102341In[4]:=
If[num/denom > Pi + E, Print[Greater], Print[Not
greater]]In[5]:= If[(num - 1)/denom > Pi + E, Print[Greater],
Print[Notgreater]]-- Clive
Toothhttp://www.clivetooth.dk===
===
Subject:
: Re: Packing squares
into a 1 x Zeta[2] rectangle>> But as a matter of interest,>>
how did you deal with pi^2 using exact arithmetic?Mathematica
does it all for me. Mathematica can do comparisons between,>
for example, rational numbers and transcendentals. Here is a
short> Mathematica session in which it compares a rational,
num/denom, with pi+e> and prints the result Greater, it then
compares the rational> (num-1)/denom with pi+e and prints the
result Not greater:In[1]:= $MaxExtraPrecision = 100> Out[1]=
100In[2]:= num =
6083355812763767537862691282377375125710417240146> Out[2]=
6083355812763767537862691282377375125710417240146In[3]:= denom
= 1038137562741240040724843633763265660486107102341> Out[3]=
1038137562741240040724843633763265660486107102341In[4]:=
If[num/denom > Pi + E, Print[Greater], Print[Not
greater]]In[5]:= If[(num - 1)/denom > Pi + E, Print[Greater],
Print[Not> greater]]I'm not entirely clear what this is
doing.using exact arithmetic (the BigInteger class).It was
actually impossible to get much past 10000as the numerators
and denominators get too big.So I wonder if Mathematica
actually works with floating-point numbersto a certain (high)
precision?One interesting point is that several other
rectangles I triedwith the same area pi^2/6 could not be
packed with these squares,eg pi/2 x pi/3 and pi/sqrt(6) x
pi/sqrt(6).In fact 1 x pi^2/6 is the only packable rectangle
of this areathat I have found, though I didn't made a serious
study.Incidentally I took pi=355/113, which I think is just
good enoughfor the range I was considering.I always took the
square 1/n x 1/n from the smallest vacant rectangle.I don't
know if another choice would be as good, or even better.The
average rectangle had been divided about 50--100 timeswith
10000 squares, which I found surprisingas the number of
remaining rectangles was (of course) a little under 10000.--
Timothy Murphy e-mail (<80k only): tim /at/
birdsnest.maths.tcd.ietel: +353-86-2336090,
+353-1-2842366s-mail: School of Mathematics, Trinity College,
Dublin 2, Ireland===
===
Subject:
: synergetic geometryCosmic fishing
this weekend in DC.http://snec.cjfearnley.com/Dick===
===
Subject:
:
Re: synergetic geometryRe: Thoughts (on geodesic-l on
listserv.acsu.buffalo.edu)I am a big fan of _S_, Bucky's opus,
but he was totally out to lunchon descartesian coordination,
denying us our modelability. I mean,wouldn't you say that it
has a lot more to dowith the lack of modelling, than the
particular coordinates,all of which are
omni-intertransformable, one to the other (or,as they say,
isometrically isomorphic) ?? just because Bucky refuzed to
learn it, relying on all sortsof student-engineers to get over
his lacunae, don'mean ****. the important theing that he did
get-- and which is introduced in Plates 1 and 2 in _S_ --is
spherical trig.; or else, he'd have lost his ship, andthus his
command in hte Navy, in teh day before radio-contact. don't
know that spherical stuff, yet?... was it because,saith Bucky,
pi=bad? as for Science Friday, that's just one of the reasons
that I stoppedlistening to the God-am radio, as with the Boob
Tube, before that;NPR's main job, now, is to assure thattwo
Bonesman vault out of a mausoleum on the Yale campus,and
hokey-dokey Copenhagen Schooler crappola just addsto the smoke
(note tahtUsama bin Laden is in a perfectly undead
superposition,with reference to the Bush and Media Hype,
althoughit wouldn't matter weather this state collapsedto dead
or alive, to Blair's McCrusade ... in the venueof
Schroedinger's joke about the Cat in the Boxwith the Atom to
Its Head .-) thus quoth:just doesn't make sense., even to the
phyicists! It seems that as we approach the limits of our
current understanding of themicro and macrocosms, many of our
concepts begin to break down. Buckycontended unflinchingly
that the problem was our use of Cartesian --Give Earth a
Trickier Dick Cheeny -- out of office, after GIGA
years.http://www.benfranklinbooks.com/http://www.rand.org/
publications/randreview/issues/rr.12.00/http://
members.tripod.com/~american_almanac===
===
Subject:
: Re: Connected
Graphs Question, Part II> A few days ago, I asked about the
percentage of all graphs that are> connected. Klaus hoffman
directed me to the onlince encyclopedia of> integere
sequences. So I calculate the ratio and I find that up till
N=20,> about 99% of graphs are connected. I calculated the
ratio of sequences> A001187 vs. A006125 and the ratio of
A001349 vs. A000088.Have I screwed up? Small graphs should be
connected but graphs with 20> nodes? It's seems that as N gets
bigger, connectedness should be an> extremely rare property. At
some point, singletons, dyads and other little> groups should
start breaking off...Someone, please enlighten me.> some
people answered you, but maybe you had something different in
mind. The problem becomes interesting if you consider graphs
with N nodes and f(N) edges and ask how fast f may grow in
order to guarantee that the frequency of connected graphs goes
to zero as N goes to infinity. I seem to recall that Erdoes and
Hajnal put up some theory on this in the 1960's .It may also be
that you are interested in *unlabelled* graphs ?! -- Stefan
Wehmeierstefanw@mupad.de===
===
Subject:
: Re: Connected Graphs
Question, Part IIHello Fabio,> Small graphs should be
connected but graphs with 20> nodes?Fred already gave a good
answer. But even the portion of graphs having adiameter of 2
tends to 100% as the size of the graph grows. This is
so,because having a diameter of two is expressible by an
extensionaxiom.A (n+m+1)-extension axiom for graphs says, that
for any two disjoinedsets N and M of vertices with |N|=n and
|M|=m, there exists a vertex qoutside of N and M, such that
edges go from q to each vertex in N andto no vertex in M. Now,
every such extension axiom is true in almostall finite graphs.
Choosing n=2 and m=0 gives you the desired property.A nice
proof is in the chapter about 0-1 laws
of:http://www.math.helsinki.fi/~logic/people/jouko.vaananen/
shortcourse.pdfGreetings,Oswald--   _/_/     _/  | _/  _/   
_/  |  Oswald Jaskolla <oswald@jaskolla.net>_/  _/  _/ _/  
|    http://www.jaskolla.net/oswald/ _/_/ _/ _/_/ _/ 
|===
===
Subject:
: Re: Connected Graphs Question, Part IIOriginator:
fab@soda.csua.berkeley.edu (Fabio Rojas)>You've got it
backwards: *disconnectedness* is a rare property for
big>graphs. Assuming that any two vertices have an independent
50% chance...>right? In other words, hardly any. So you
shouldn't expect very many>singletons to break off from a
really big random graph.Hmmm... can't argue with that logic. I
just imagined the creation of longpaths as unlikely... the
matching up of arcs for a path between all pairsGrumble... now
I gotta go and debug a program... should've taken some graph
theory... Grrr..fabio===
===
Subject:
: Re: Connected Graphs
Question, Part II
3QLpj-NoP*NzsIC,boYU]bQ]H'<P%JD/,.J>y<#4ga3$21:>You've got it
backwards: *disconnectedness* is a rare property for
big>graphs. Assuming that any two vertices have an independent
50% chance> ...>right? In other words, hardly any. So you
shouldn't expect very many>singletons to break off from a
really big random graph.Hmmm... can't argue with that logic. I
just imagined the creation of long> paths as unlikely... the
matching up of arcs for a path between all pairs> Grumble...
now I gotta go and debug a program... should've taken > some
graph theory... Grrr..The paths don't have to be long.If each
edge has an independent 50% chance of existing, then the
probability that there is no two-edge path joining two
vertices x,y is (3/4)^{n-2} (there are n-2 choices for the
intermediate vertex and each choice involves unrelated edges).
Therefore, the probability that there exists any pair x,y
without a two-edge path is at most n(n-1)/2 * (3/4)^{n-2} (at
most rather than equal because the probabilities are not
independent for different pairs x,y). But if all pairs have
such a graph then the graph is connected.So, with probability
at least 1 - n(n-1)(3/4)^{n-2}/2 the graph is connected. This
probability goes to 1 very quickly in the limit of large n.--
David Eppstein http://www.ics.uci.edu/~eppstein/Univ. of
California, Irvine, School of Information & Computer
Science===
===
Subject:
: Re: Connected Graphs Question, Part
II>You've got it backwards: *disconnectedness* is a rare
property for big>graphs.Let G be a connected graph of order n.
If any one vertex isdisconnected, the graph is disconnected,
right. So theoreticallythere would be at least |n|
disconnected graphs for every connectedgraph.Of course, one
may take a disconnected graph and create numberousconnected
graphs.So where are we? Back to randomly generated graphs?
Probably!===
===
Subject:
: Re: Connected Graphs Question, Part II
3QLpj-NoP*NzsIC,boYU]bQ]H'<P%JD/,.J>y<#4ga3$21:> Let G be a
connected graph of order n. If any one vertex is>
disconnected, the graph is disconnected, right. So
theoretically> there would be at least |n| disconnected graphs
for every connected> graph.Um. There would be, say, at least n
disconnected graphs on n vertices for every connected graph on
n-1 vertices (one for each way of disconnecting a singleton and
connecting the rest of the vertices, assuming the vertices are
labeled). But the number of n-vertex graphs is a factor of
2^{n-1} larger than the number of (n-1)-vertex graphs (there
are n-1 more edges that each may or may not exist).This factor
of 2^{n-1} swamps the factor of n coming from your argument.--
David Eppstein http://www.ics.uci.edu/~eppstein/Univ. of
California, Irvine, School of Information & Computer
Science===
===
Subject:
: Re: Connected Graphs Question, Part II>>
Have I screwed up? Small graphs should be connected but graphs
with>> 20 nodes? It's seems that as N gets bigger,
connectedness should be>> an extremely rare property. At some
point, singletons, dyads and>> other little groups should
start breaking off...>You've got it backwards:
*disconnectedness* is a rare property for big>graphs. Assuming
that any two vertices have an independent 50% chance>of being
joined by an edge, Or better yet, note that if we consider a
uniform distribution on allgraphs with N nodes then any two
vertices _do_ have a 50% chanceof being joined by an edge, and
that these events _are_ independentfor distinct pairs of nodes.
(Heh, was contemplating how good amodel your assume
gave...)>what's the probability that a given node>in a 20-node
graph will be isolated? (.5)^19, right? So how many of>the 20
nodes would you expect to be isolated, on average?
20*(.5)^19,>right? In other words, hardly any. So you
shouldn't expect very many>singletons to break off from a
really big random graph.===
===
Subject:
: Re: Well-order the reals
(was GCDs of Infinite Set of Integer Pairs)
<slrnc4q12e.ria.willem@toad.stack.nl> <c2jf0b$sil$1@m
<Pine.LNX.4.58-035.0403122139540.32077@unix43..cmu.edu> Which
reminds me of something tangentially related to where>>this
mess has now drifted: I've heard that you can well-order>>the
set of real numbers [such that every set of real numbers>>will
have a unique least member under this ordering]. How?>>(Or is
this one of those things that has been proven
deductively,>>but never by example?)Try this example out.
Well-ordering is a new concept to me but it seemslike all you
need is a function which will determine the least member of
anysubset. Here's one, in algorithmic rule form (first rule to
fire determinesthe result).least(a,b) if |a| > |b| return b if
|a| < |b| return a if a < 0 return a return bSo in the set of
all reals, 0 is the least element, and the rest are sortedby
absolute value and then by sign. Does that make sense? Is it a
goodexample of well-ordering?--Keith Lewis klewis {at}
mitre.orgThe above may not (yet) represent the opinions of my
employer.===
===
Subject:
: Re: Well-order the reals (was GCDs of
Infinite Set of Integer Pairs) Which reminds me of something
tangentially related to where>>this mess has now drifted: I've
heard that you can well-order>>the set of real numbers [such
that every set of real numbers>>will have a unique least
member under this ordering]. How?>>(Or is this one of those
things that has been proven deductively,>>but never by
example?)Try this example out. Well-ordering is a new concept
to me but it seems> like all you need is a function which will
determine the least member of any> subset. Here's one, in
algorithmic rule form (first rule to fire determines> the
result).least(a,b)> if |a| > |b| return b> if |a| < |b| return
a> if a < 0 return a> return bSo in the set of all reals, 0 is
the least element, and the rest are sorted> by absolute value
and then by sign. Does that make sense? Is it a good> example
of well-ordering?--Keith Lewis klewis {at} mitre.org> The
above may not (yet) represent the opinions of my employer.For
a set, S, to be well ordered, every non-empty subset must have
as a member a first or smallest element.The set, T, of reals
greater than zero is not well ordered by your ordering: given
x in T, then x/2 < x and x/2 is also in T, so there can be no
smallest element of T.If one accepts the axiom of choice, the
reals must have a well ordering, but, as far as I know, no one
has come up with an explicit example of one.===
===
Subject:
: Re:
Well-order the reals (was GCDs of Infinite Set of Integer
Pairs) <Pine.LNX.4.58-035.0403122139540.32077@unix43..cmu.edu>
<c34j7b$2d$1@newslocal.mitre.org> Which reminds me of something
tangentially related to where>this mess has now drifted: I've
heard that you can well-order>the set of real numbers [such
that every set of real numbers>will have a unique least member
under this ordering]. How?>(Or is this one of those things that
has been proven deductively,>but never by example?)>Try this
example out. Well-ordering is a new concept to me but it
seems>like all you need is a function which will determine the
least member of any>subset. Here's one, in algorithmic rule
form (first rule to fire determines>the result).>least(a,b)>
if |a| > |b| return b> if |a| < |b| return a> if a < 0 return
a> return bThis gives you a total ordering, but does not give
you a wellordering.One can define a total ordering is a
partial ordering in which anysubset with exactly two elements
has a least element. A wellordering is a generalization that
asks that every ->nonempty<- subsethave a least element. The
two-element case deals with all finite casesby induction, so
well ordering starts being interesting when youhave infinite
subsets.So, for example, take your proposed ordering (which is
a perfectlyfine total ordering). What is the least element of{x
in R: 0<|x|<=1}, a nonempty subset?>So in the set of all reals,
0 is the least element, and the rest are sorted>by absolute
value and then by sign. Does that make sense? Is it a
good>example of well-ordering?No. Because it is not a
well-ordering, it is only a total ordering.(Don't scratch your
head too much: you cannot exhibit a well-orderingof the reals
in the sense that you need to invoke the Axiom of Choiceto
prove there is one; they are non-constructible)--
==============================================================
========It's not denial. I'm just very selective about what I
accept as reality. --- Calvin (Calvin and
Hobbes)=======================================================
==================
===
Subject:
: Re: Well-order the reals (was GCDs
of Infinite Set of Integer Pairs)) Try this example out.
Well-ordering is a new concept to me but it seems) like all
you need is a function which will determine the least member
of any) subset. Here's one, in algorithmic rule form (first
rule to fire determines) the result).)) least(a,b)) if |a| >
|b| return b) if |a| < |b| return a) if a < 0 return a) return
b)) So in the set of all reals, 0 is the least element, and the
rest are sorted) by absolute value and then by sign. Does that
make sense? Is it a good) example of well-ordering?Nope. Only
works on finite subsets.What's the least member of the
set:'all reals between, but not including, 0 and 1' ?SaSW,
Willem-- Disclaimer: I am in no way responsible for any of the
statements made in the above text. For all I know I might be
drugged or something.. No I'm not paranoid. You all think I'm
paranoid, don't you !#EOT===
===
Subject:
: Re: Well-order the reals
(was GCDs of Infinite Set of Integer Pairs)>Try this example
out. Well-ordering is a new concept to me but it seems>like
all you need is a function which will determine the least
member of any>subset.Constructing a well-ordering for the
reals requires the Axiom ofChoice, which is independent from
all the other axioms of set theory,so any construction can not
use any of the other axioms.Which makes it a bit difficult.--
I'm not interested in mathematics that might have anythingto
do with reality. -- Easterly, in sci.math===
===
Subject:
: Re:
Well-order the reals (was GCDs of Infinite Set of Integer
Pairs)>>Try this example out. Well-ordering is a new concept
to me but it seems>>like all you need is a function which will
determine the least member of any>>subset.> Constructing a
well-ordering for the reals requires the Axiom of> Choice,
which is independent from all the other axioms of set theory,>
so any construction can not use any of the other axioms.Are you
saying that AC is equivalent to the existence of a
well-orderingof the reals?To put the question another way, are
you saying that a well-orderingof one particular set (the
reals) implies that *every* set can be wellordered? As far as
I am aware, there is no proof of that.> Which makes it a bit
difficult.Yes.-- Dave SeamanJudge Yohn's mistakes revealed in
Mumia Abu-Jamal
ruling.<http://www.commoncouragepress.com/index.cfm?action=
book&bookid=228>===
===
Subject:
: Re: Well-order the reals (was GCDs
of Infinite Set of Integer Pairs) Adjunct Assistant Professor
at the University of Montana.>Try this example out.
Well-ordering is a new concept to me but it seems>like all you
need is a function which will determine the least member of
any>subset.>> Constructing a well-ordering for the reals
requires the Axiom of>> Choice, which is independent from all
the other axioms of set theory,>> so any construction can not
use any of the other axioms.>Are you saying that AC is
equivalent to the existence of a well-ordering>of the
reals?No: but you need AC up to a certain cardinal (either c
or 2^c, can'tremember right now), which is equivalent to
well-ordering thereals. AC is equivalent to the Well Ordering
Theorem (every set can bewell ordered).--
==============================================================
========It's not denial. I'm just very selective about what I
accept as reality. --- Calvin (Calvin and
Hobbes)=======================================================
==================
===
Subject:
: Re: Well-order the reals (was GCDs
of Infinite Set of Integer Pairs)>Are you saying that AC is
equivalent to the existence of a well-ordering>of the
reals?No.>To put the question another way, are you saying that
a well-ordering>of one particular set (the reals) implies that
*every* set can be well>ordered? As far as I am aware, there
is no proof of that.Most likely not, but I doubt very much
even the reals can bewell-ordered in any meaningful way. Who
knows about greatercardinalities.But were it possible to
well-order a set of cardinality c, thenintuitively it would
make sense that we could well-order any set sincewe would have
a way of well-ordering the power set of the naturals.The same
method could then be used to well-order the power set of
thecontinuum, etc.But this is of course theoretical, or should
I say theologicalpondering, not proof.-- I'm not interested in
mathematics that might have anythingto do with reality. --
Easterly, in sci.math===
===
Subject:
: Re: Well-order the reals (was
GCDs of Infinite Set of Integer Pairs)>Are you saying that AC
is equivalent to the existence of a well-ordering>of the
reals?No.>To put the question another way, are you saying
that a well-ordering>of one particular set (the reals) implies
that *every* set can be well>ordered? As far as I am aware,
there is no proof of that.Most likely not, but I doubt very
much even the reals can be> well-ordered in any meaningful
way. Who knows about greater> cardinalities.>
[...snipped]There's an interesting result (which I need to
think about to recallexactly) that there is no way to define a
well ordering of the reals inany way that we would consider
constructive. I interpret this as saying that accepting the
axiom of choice is darnsilly since it leads us to the
conclusion that the reals are wellorderable, but there is no
way for me see what it is!===
===
Subject:
: Re: Well-order the reals
(was GCDs of Infinite Set of Integer Pairs) Which reminds me of
something tangentially related to where>this mess has now
drifted: I've heard that you can well-order>the set of real
numbers [such that every set of real numbers>will have a
unique least member under this ordering]. How?>(Or is this one
of those things that has been proven deductively,>but never by
example?)Try this example out. Well-ordering is a new concept
to me but it seems> like all you need is a function which will
determine the least member of> any> subset. Here's one, in
algorithmic rule form (first rule to fire> determines the
result).least(a,b)> if |a| > |b| return b> if |a| < |b| return
a> if a < 0 return a> return bSo in the set of all reals, 0 is
the least element, and the rest are> sorted> by absolute value
and then by sign. Does that make sense? Is it a good> example
of well-ordering?It isn't a well-ordering. The set {x in R: x
> 0} has no least elementin this ordering.-- Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92;
mistakes his penis for a square root, 88-9Francis Wheen, _How
Mumbo-Jumbo Conquered the World_===
===
Subject:
: Re: Well-order the
reals (was GCDs of Infinite Set of Integer Pairs)>>
least(a,b)>> if |a| > |b| return b>> if |a| < |b| return a>>
if a < 0 return a>> return b>>> So in the set of all reals,
0 is the least element, and the rest are>> sorted>> by
absolute value and then by sign. Does that make sense? Is it a
good>> example of well-ordering?>It isn't a well-ordering. The
set {x in R: x > 0} has no least element>in this ordering.Doh!
--Keith Lewis klewis {at} mitre.orgThe above may not (yet)
represent the opinions of my employer.===
===
Subject:
: Re: Formula
for Pi by support1.mathforum.org (8.11.6/8.11.6/The Math
Forum, $Revision: 1.9 primary) id i2FFefd00981;>> Formula for
Pi>>> Several years ago, while at high school, I derived the
following formula for Pi:>>> Pi = lim 2^(n+1)*sqrt(2 - sqrt(2
+ sqrt(2 + sqrt(2 + ...)))).>> n->infinity <----------- n times
----------->> The beauty of this formula, I think, is that it
can be derived using only Pythagoras formula.>>> Now I wonder
if it already exists in the mathematical literature. Probably
it does. If>> affirmative, I would appreciate any somebody
telling me where.>Your formula is formula (44) from><a
href=http://mathworld.wolfram.com/PiFormulas.html>http://
mathworld.wolfram.com/PiFormulas.html</aBest regards,>Jose
Carlos Santos===
===
Subject:
: Re: Games/Puzzles With Advanced
Mathematics?>what are some good examples of puzzles and games
which use *advanced*>mathematics?>I wonder specifically here
about easy-to-understand puzzles/games>where the trick to
solving the puzzle or successfully playing the>game uses some
math way beyond most people, and somewhat (at least)>beyond
most mathematicians.Bcing easy to understand against advanced
leaves evaluatinganswers somewhat ambiguous. Here are two
nominations:Blackjack (21) with Kelly Betting, derived from
Claude Shannon'sinformation theory producing consistent and
assured wins (whencombined with an ability to count cards)
:http://bjmath.com/bjmath/kelly/kelly.pdfAir-to-air combat
with differential games as the source of the mathrequired to
express the
problemshttp://faculty.gvsu.edu/aboufade/web/szurley.htm John
Baileyhttp://home.rochester.rr.com/jbxroads/mailto.html===

===
Subject:
: Re: Games/Puzzles With Advanced Mathematics?> I
wonder specifically here about easy-to-understand
puzzles/games> where the trick to solving the puzzle or
successfully playing the> game uses some math way beyond most
people,The usual example here is the game of Nim.-- G. A.
Edgar http://www.math.ohio-state.edu/~edgar/===
===
Subject:
: Re:
Games/Puzzles With Advanced Mathematics?> I am just curious,>
what are some good examples of puzzles and games which use
*advanced*> mathematics?The Mecca of prime puzzles,
conjectures and problmes of any difficultyis :
http://www.primepuzzles.netof Carlos Rivera. L.R===
===
Subject:
:
Re: Games/Puzzles With Advanced Mathematics?> I am just
curious,> what are some good examples of puzzles and games
which use *advanced*> mathematics?The Mecca of prime puzzles,
conjectures and problmes of any difficulty> is :
http://www.primepuzzles.net> of Carlos Rivera. L.RTo be
honest, apart from the occasional puzzle that touches on old
unsolved conjectures (those of Erdos etc.), the mathematics on
thatsite is not advanced.Typically the only tools you need are
Fermat's little Theorem, the Chinese Remainder Theorem, and
access to something like Pari/GP, UBasic, Mathematica, etc.
for the grunt-work. That's not to denigrate the site in any
way. I love it, it's right up my street, but I'm not an
advanced mathematician in any way.Primepuzzles.net is more of
an exercise of algorithmics, and the difference in rate of
finding answers can be unimaginably vast in some situations
(e.g. the minimal titanics for puzzles 257 and 258 can only be
found using the right algorithm, which uses no advanced maths
at all).-- 1st bug in MS win2k source code found after 20
minutes: scanline.cpp2nd and 3rd bug found after 10 more
minutes: gethost.cBoth non-exploitable. (The 2nd/3rd ones
might be, depending on the CRTL)===
===
Subject:
: Re: Games/Puzzles
With Advanced Mathematics?> I wonder specifically here about
easy-to-understand puzzles/games> where the trick to solving
the puzzle or successfully playing the> game uses some math
way beyond most people, and somewhat (at least)> beyond most
mathematicians.> Leroy QuetI have a game with primes where the
solution of some its questionsrequire searching in mathematical
Internet pages, (Example, that of Tony Forbes) for its
answer.It is related with arithmetical progresions, cluster of
primes andother patterns.If someone is interested I can send
free my program: STARRY.EXELuis Rodriguez===
===
Subject:
: Re:
Games/Puzzles With Advanced Mathematics?>> I wonder
specifically here about easy-to-understand puzzles/games
where>> the trick to solving the puzzle or successfully
playing the game uses>> some math way beyond most people, and
somewhat (at least) beyond most>> mathematicians.Rubiks' Cube;
group theory.I'm not sure about that. I've seen group theory
used to estimate how manyconfigurations are possible, and to
estimate the maximum number of moves acube can be away from
solved, and stuff like that, but I've not seenanything where
group theory was actually used *solve* the damned thing!About
the only difference I've seen between people who know group
theory andpeople who don't when it comes to solving Rubik's
cubes is that when youhave some transform that does something
useful, say rotates two adjacentcorners, and you need to do
that but the things you need to affect are inthe wrong
position (e.g, opposite corners instead of adjacent), and so
youdo a transform to get the things into position, apply your
transform, andthen invert the transform that got them into
position, is that the grouptheory people know that they are
using conjugacy, whereas the others justthink of it as doing
something obvious. :-)Can one actually use group theory to
solve the cube?-- --Tim Smith===
===
Subject:
: Re: Games/Puzzles
With Advanced Mathematics?Mime-version: 1.0Content-type:
text/plain; charset=US-ASCIIContent-transfer-encoding: 7bit> I
wonder specifically here about easy-to-understand puzzles/games
where> the trick to solving the puzzle or successfully playing
the game uses> some math way beyond most people, and somewhat
(at least) beyond most> mathematicians.and I offered:>>
Rubiks' Cube; group theory.Tim Smith> I'm not sure about that.
...Well, OK, I disagree with you then. But rather than argue
it, I'll offer upDots & Boxes.Bob H===
===
Subject:
: Re:
Games/Puzzles With Advanced Mathematics?Harris>> I wonder
specifically here about easy-to-understand puzzles/games
where>> the trick to solving the puzzle or successfully
playing the game uses>> some math way beyond most people, and
somewhat (at least) beyond most>> mathematicians. Rubiks'
Cube; group theory.> I'm not sure about that. I've seen group
theory used to estimate how many> configurations are possible,
and to estimate the maximum number of moves a> cube can be away
from solved, and stuff like that, but I've not seen> anything
where group theory was actually used *solve* the damned
thing!> About the only difference I've seen between people who
know group theoryand> people who don't when it comes to solving
Rubik's cubes is that when you> have some transform that does
something useful, say rotates two adjacent> corners, and you
need to do that but the things you need to affect are in> the
wrong position (e.g, opposite corners instead of adjacent),
and so you> do a transform to get the things into position,
apply your transform, and> then invert the transform that got
them into position, is that the group> theory people know that
they are using conjugacy, whereas the others just> think of it
as doing something obvious. :-)> Can one actually use group
theory to solve the cube?Yes.===
===
Subject:
: Re: Games/Puzzles
With Advanced Mathematics?> I wonder specifically here about
easy-to-understand puzzles/games where> the trick to solving
the puzzle or successfully playing the game uses> some math
way beyond most people, and somewhat (at least) beyond most>
mathematicians.>>> Rubiks' Cube; group theory.I'm not sure
about that. I've seen group theory used to estimate how many>
configurations are possible, and to estimate the maximum
number of moves a> cube can be away from solved, and stuff
like that, but I've not seen> anything where group theory was
actually used *solve* the damned thing!About the only
difference I've seen between people who know group theory> and
people who don't when it comes to solving Rubik's cubes is that
when> you have some transform that does something useful, say
rotates two> adjacent corners, and you need to do that but the
things you need to> affect are in the wrong position (e.g,
opposite corners instead of> adjacent), and so you do a
transform to get the things into position,> apply your
transform, and then invert the transform that got them into>
position, is that the group theory people know that they are
using> conjugacy, whereas the others just think of it as doing
something obvious.> :-)Can one actually use group theory to
solve the cube?> Peter Neumann is said to have done so before
even seeing onein the flesh :-)-- Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92;
mistakes his penis for a square root, 88-9Francis Wheen, _How
Mumbo-Jumbo Conquered the World_===
===
Subject:
: Re: Games/Puzzles
With Advanced Mathematics?> I am just curious, what are some
good examples of puzzles and games which> use *advanced*
mathematics?What advanced mathematics means exactly here is
subjective.This might be one. I'm not sure, because I don't
know the solution. I dorecall that when it was published in a
magazine, they only received onecorrect solution, and a bunch
of incorrect solutions, and the correctsolution was supposedly
not elementary.Sorry for the vagueness of this, but I'm going
from memory here. You have a set of people randomly scattered
over an area. (I do not remember if it was N people over a
particularly shaped region, or an infinite number of people
over a plane at a specified density). Each has a gun. At a
specified time, each one shoots his nearest neighbor. What is
the probability that a given person is not shot?I think the
magazine was Omni.-- --Tim Smith===
===
Subject:
: Re: Games/Puzzles
With Advanced Mathematics?>I am just curious,>what are some
good examples of puzzles and games which use
*advanced*>mathematics?You need to read rec.games.abstract for
a while!>What advanced mathematics means exactly here is
subjective.>I am more interested in examples of games and
puzzles for the general>audience.The card game 'Set' is all
about finding lines in a 4-dimensional affinespace over the
Galois field of 3 elements. Yet five-year-olds can play!>I am
aware that some advanced math has been used to analyze
some>famous games, for example. (Although precise examples
slip my mind>now.)When Internet flotsam on these games comes
my way I try to save a noteabout it at
http://www.math.niu.edu/~rusin/uses-math/games/>Leroy
QuetLeroy, how many q's are we supposed to remove from the
email addressto get through to you privately? My attempts
failed. (Did I removethe wrong q ?)dave===
===
Subject:
: Re:
Games/Puzzles With Advanced Mathematics?>I am just
curious,>what are some good examples of puzzles and games
which use *advanced*>mathematics?You need to read
rec.games.abstract for a while!>What advanced mathematics
means exactly here is subjective.I am more interested in
examples of games and puzzles for the general>audience.The
card game 'Set' is all about finding lines in a 4-dimensional
affine> space over the Galois field of 3 elements. Yet
five-year-olds can play!>I am aware that some advanced math
has been used to analyze some>famous games, for example.
(Although precise examples slip my mind>now.)When Internet
flotsam on these games comes my way I try to save a note>
about it at http://www.math.niu.edu/~rusin/uses-math/games/Leroy QuetLeroy, how many q's are we supposed to remove from
the email address> to get through to you privately? My
attempts failed. (Did I remove> the wrong q ?)daveI hope you
have seen a few of my games I have posted to sci.math
andrec.puzzles.:)(Most being played on a grid or number-line.
Just do a search forLeroy Quet and game.)As for my email,
there is a really hyphen between the second and thirdq.(I hope
spammers cannot somday soon figure out, with AI-bots, what
Imean here...But they WILL break all our human-only-readable
spam-blockseventually...):(Leroy Quet===
===
Subject:
: Re:
Games/Puzzles With Advanced Mathematics?> I am just curious,>
what are some good examples of puzzles and games which use
*advanced*> mathematics?> What advanced mathematics means
exactly here is subjective.> I am more interested in examples
of games and puzzles for the general> audience.> I am aware
that some advanced math has been used to analyze some> famous
games, for example. (Although precise examples slip my mind>
now.)> I wonder specifically here about easy-to-understand
puzzles/games> where the trick to solving the puzzle or
successfully playing the> game uses some math way beyond most
people, and somewhat (at least)> beyond most
mathematicians.There was a logic game that used to be
advertised in science magazines -Wiff'n'proof or something
like that.===
===
Subject:
: Re: Games/Puzzles With Advanced
Mathematics?Richard Henry:> There was a logic game that used
to be advertised in science magazines -> Wiff'n'proof or
something like that.WFF'n'Proof, where WFF stood for
Well-Formed Formula and meant anexpression in Boolean algebra
using a prefix (Lukasiewicz or Polish)notation. There were
special dice with the characters needed toconstruct these
expressions. The game was really a series ofeducational games
of progressively greater complexity, and the finalones in the
series (which I never played) involved actually
constructingproofs, or at least making deductions, using this
notation.The name was presumably a pun on Whiffenpoof.-- Mark
Brader, Toronto | Professor, I think I have a
counterexample.msb@vex.net | That's all right; I have two
proofs.===
===
Subject:
: Re: Games/Puzzles With Advanced
Mathematics?> I am just curious,> what are some good examples
of puzzles and games which use *advanced*> mathematics?What
advanced mathematics means exactly here is subjective.I am
more interested in examples of games and puzzles for the
general> audience.I am aware that some advanced math has been
used to analyze some> famous games, for example. (Although
precise examples slip my mind> now.)I wonder specifically here
about easy-to-understand puzzles/games> where the trick to
solving the puzzle or successfully playing the> game uses some
math way beyond most people, and somewhat (at least)> beyond
most mathematicians.Leroy QuetIf you check out the Clay
Mathematics Institute's Millineum PrizeProblems, you'll see
the P vs. NP problem has been closely associatedwith the game
Minesweeper.http://www.claymath.org/Millennium_Prize_Problems/
P_vs_NP/Cheers,Bobby===
===
Subject:
: Re: Games/Puzzles With
Advanced Mathematics?>If you check out the Clay Mathematics
Institute's Millineum Prize>Problems, you'll see the P vs. NP
problem has been closely associated>with the game
Minesweeper.>http://www.claymath.org/Millennium_Prize_Problems
/P_vs_NP/That's a rather curious way of putting it.
Minesweeper is only one example of an NP-complete problem. So
is Go, and so is a suitablygeneralized version of chess
IIRC.Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada V6T
1Z2===
===
Subject:
: Re: Games/Puzzles With Advanced
Mathematics?Leroy Quet> I am just curious,> what are some good
examples of puzzles and games which use *advanced*>
mathematics?> What advanced mathematics means exactly here is
subjective.> I am more interested in examples of games and
puzzles for the general> audience.> I am aware that some
advanced math has been used to analyze some> famous games, for
example. (Although precise examples slip my mind> now.)> I
wonder specifically here about easy-to-understand
puzzles/games> where the trick to solving the puzzle or
successfully playing the> game uses some math way beyond most
people, and somewhat (at least)> beyond most
mathematicians.The topological game called Hex, invented by
the poet-architect Piet Hein(and independantly by the famous
John Nash) is poorly understood, althoughpurely
mathematical.LH===
===
Subject:
: Re: Games/Puzzles With Advanced
Mathematics?>I am just curious,>what are some good examples of
puzzles and games which use *advanced*>mathematics?>What
advanced mathematics means exactly here is subjective.>I am
more interested in examples of games and puzzles for the
general>audience.>I am aware that some advanced math has been
used to analyze some>famous games, for example. (Although
precise examples slip my mind>now.)>I wonder specifically here
about easy-to-understand puzzles/games>where the trick to
solving the puzzle or successfully playing the>game uses some
math way beyond most people, and somewhat (at least)>beyond
most mathematicians.>Leroy QuetSolving the Eternity puzzle was
way beyond most mathematicians. Not sure if thisis the sort of
'math' you wanted though :>-- Patrick Hamlyn posting from
Perth, Western AustraliaWindsurfing capital of the Southern
HemisphereModerator: polyforms group
(polyforms-subscribe@egroups.com)===
===
Subject:
: Re: Pi and Po>
Given Pi, is there a formula for calculating the n'th digit of
Pi that does> NOT use floors and ceilings? Yes, See:
http://crd.lbl.gov/~dhbailey/===
===
Subject:
: Re: Hints for self
taught topology?Originator: baez@math-cl-n03.math.ucr.edu
(John Baez)>| Are there any common pitfalls, etc. that I need
to be ready for,>|or anything else I should know for this
mission?>probably the most important thing you should know is
that galois>theory and the so-called algebraic topology part
of topology (which>is a big part of what munkres's text is
about) are, in a sense,>secretly the same
subject.Unfortunately you can only figure out in *what* sense
by eitherthinking *really* hard, asking someone to explain it
to you, orfinding the right book. I think Michio Kuga's
Galois' Dream: Group Theory and Differential Equations might
be the most elementarybook that comes close to explaining this
stuff. In practice you may need to knock your head on algebraic
topologyand Galois theory a little before you're ready to
profit from Dolan'sadvice, though in some alternate universe
there could be some reallyfun easy textbook that would explain
algebraic topology and Galoistheory simultaneously from
scratch, making it clear how they're flipsides of the same
thing.It can't hurt to know right from the start that homology
theoryis like a watered-down version of homotopy theory,
abelian andeasier to calculate with, but less conceptual -
until you reach aconceptual understanding of what this
watering-down process reallyis. I'm sure you'll find these
cryptic remarks extremely useful. :-)Here's a good free online
introduction to algebraic
topology:http://www.math.cornell.edu/~hatcher/AT/ATpage.html==
=
===
Subject:
: Re: 1st-order Pigeonhole Challenge : : > : > Well,
actually, he was full of . : > : Nonsense.And you're full of
too. : that straight, I can't help.Dip, I ALREADY MADE AND
REBUTTED HIS FULL QUOTE.Jeezus. : [quote] :
------------------------------------------------- : : (i)
There are at least m objects : (ii) There are exactly n A's :
(iii) The f of each object is an A : Therefore, : (iv) There
are two distinct objects a, b such that f(a) = f(b). : : For
any particular choices of m and n, we can express (i)-(iv) as
: first-order formulas (note that m and n are not variables).
: : When we express (i) as a first-order formula, it just
contains a : sequence of m existential quantifiers, followed
by conjunction of : inequalities between the variables). : :
If m > n, this argument is valid in FOL. : :
------------------------------------------------- : [/quote] :
: Indeed: : : For any particular choices of m and n, we
***can*** : express (i)-(iv) as first-order formulas.Not
without attributing something entirely mystical to =.= is
irrelevant to this problem IF you already have <, which isin
fact MORE indispenbsable. More to the point,this whole thing
is doable AT 0TH-ORDER, WHICH MAKES EVERYthing you are saying
IRRELEVANT, AS it waswhen he said it the first
time.===
===
Subject:
: The Saga of the Dripping Stalactite... ...
has been newly uploaded as GuruGram $35
ashttp://www.tinaja.com/glib/stalac.pdfIt celebrates the
beauty of applied math, especially when exploring longstrings
of strange numbers.Additional GuruGrams at
http://www.tinaja.com/gurgrm01.aspPlease also see
http://www.tinaja.com/info01.asp
andhttp://www.tinaja.com/advt01.asp-- Don LancasterSynergetics
3860 West First Street Box 809 Thatcher, AZ 85552voice:
(928)428-4073 email: don@tinaja.com Please visit my GURU's
LAIR web site at http://www.tinaja.com===
===
Subject:
: Re: The Saga
of the Dripping Stalactite (magic sine waves)Don Lancaster
<don@tinaja.com> says...>[The Saga of the Dripping
Stalactite]> has been newly uploaded as GuruGram $35
as>http://www.tinaja.com/glib/stalac.pdfThis is an important
advance in the area of magic sine waves.Possible future
directions this research could take:...Further exploration of
the math theory behind the results....A (distributed?) brute
force search for even better solutions using a more efficient
language than the Postscript/Javascript that Don is
using....Efficient implementations on common
microcontrollers.-- Guy Macon, Electronics Engineer & Project
Manager for hire. Remember Doc Brown from the _Back to the
Future_ movies? Do you have an impossible engineering project
that only someone like Doc Brown can solve? My resume is at
http://www.guymacon.com/ ===
===
Subject:
: Size Frequency
Distribution of Large Impact Craters by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
i2FHFXj13297;I would like to understand how to use the
Hartmann Production Function to calculate the probability that
a meteorite of a given diameter will strike the surface of the
Moon. I would also like to use this function to get the
absolute number of meteorites of a given diameter, if
possible.I know what integrals are but my calculus is weak. I
am also not sure if I need to know the number of meteorites of
a given diameter. Here is the function and the coefficient
table. Maybe someone could run through one calculation for me,
or solve the equation for fake values.log10(N) = a0 + lim(n = 1
to 11 ) an[log10(D)]^nTABLE 1. Coefficients in equation (2).Old
N(D) New N(D) New N(D) R(D) for Projectilesai (Neukum, 1983)
(Neukum et al., 2001) Sensibility* (Ivanov et al., 2001)a0
3.0768 3.0876 a1 3.6269 3.557528 3.8% +1.375a2 +0.4366
+0.781027 3.9% +0.1272a3 +0.7935 +1.021521 2.5% 1.2821a4
+0.0865 0.156012 1.6% 0.3075a5 0.2649 0.444058 0.88% +0.4149a6
0.0664 +0.019977 1.3% +0.1911a7 +0.0379 +0.086850 0.78%
0.04261a8 +0.0106 0.005874 1.8% 0.03976a9 0.0022 0.006809 1.8%
3.1802 [Times] 103a10 5.18 [Times] 104 +8.25 [Times] 10 4 5.6%
+2.799 [Times] 103a11 +3.97 [Times] 105 +5.54 [Times] 10 5
24.1%+6.892 [Times] 104a12    +2.614 [Times] 106a13    1.416
[Times] 105a14    1.191 [Times] 106*Sensibility is the
coefficient variation that changes the N(D) value a factor of
2 up and down.===
===
Subject:
: Is this problem NP-Hard?Let V =
{v_0, v_1, ...} be a set of boolean variables and M a boolean
matrixwhere each element e_ij represents the value of the
variable v_i at time j.Find all the subsets S_s = {s_st | s_st
in V} with cardinality k > N suchthat s_s0 + s_s1 + ... = TRUE
for every t.===
===
Subject:
: Re: algebra homomorphisms> When does a
k-algebra homomorphism give rise to a ring
homomorphism.k-algebra homomorphisms are always ring
homomorphisms.-- Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92;
mistakes his penis for a square root, 88-9Francis Wheen, _How
Mumbo-Jumbo Conquered the World_===
===
Subject:
: Re: Cantor
Paradox>> Cantor's diagonal construction is a challenge.>> He
says to all the world, in effect, give me any list of reals,>>
numbered by the naturals, and I will show that there is a real
number>> not in the list.Yes.> Nathan has has given a list of
reals, numbered by the naturals.> Cantor's argument shows
there is a real not in the list.>> In fact, a minor
variation on his construction will easily find>> countably
many numbers not in the list, so that there are at least as>>
many numbers not in the list as in the list.>> Until you
produce such a list, you have not challenged him.I am not the
one claiming the definable numbers are countable.> Others have
claimed that Cantor's number is not definable.There is no
Cantor's number.-- Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.htmlLacan, Jacques, 79, 91-92;
mistakes his penis for a square root, 88-9Francis Wheen, _How
Mumbo-Jumbo Conquered the World_===
===
Subject:
: Re: Cantor
Paradox<logiclab@comcast.net>:> I am not the one claiming the
definable numbers are countable.> Others have claimed that
Cantor's number is not definable.> I think these people are
shooting themselves in the foot.> If Cantor's missing number
is undefinable in this> situation, how can it be definable in
any situation?Strange that you think there is such a thing as
Cantor's number.Given a list, Cantor's construction produces a
number not on thatlist. Given two different lists, it produces
two diferent numbers.So the nearest you can get to somthing
you might call Cantor'snumber is Cantor's number constructed
from a given list. Given the list, that number is definable.
It also isn't on the list. So if we have a recursive
enumeration K of the definable numbers, wecan construct a
number (let's call it C(K) )by applying Cantor'sargumnet to a
rgument to K and that number is pretty obviouslydefinable.
Hence K clearly can't exist, else we would have acontradiction
(a definable number C(K) which isn't in the list K ofall
definable numbers). So we can conclude that the set of
definablenumbers is not recursively enumerable. That doesn't
tell us anythingabout whether or not it is countable - there
are plenty of sets (atleast in a non-constructive universe)
which are countable that are notrecursively enumerable. M.[my
real email address has no no in it]===
===
Subject:
: Re: Cantor
Paradox> Cantor's diagonal construction is a challenge. He
says to all the world, in effect, give me any list of reals,>
numbered by the naturals, and I will show that there is a real
number> not in the list.Yes.> Nathan has has given a list of
reals, numbered by the naturals.> Cantor's argument shows
there is a real not in the list.In fact, a minor variation on
his construction will easily find> countably many numbers not
in the list, so that there are at least as> many numbers not
in the list as in the list. Until you produce such a list, you
have not challenged him.I am not the one claiming the definable
numbers are countable.> Others have claimed that Cantor's
number is not definable.> I think these people are shooting
themselves in the foot.> If Cantor's missing number is
undefinable in this> situation, how can it be definable in any
situation?The numbers definable in a specific language L are
countable. IfNathan's list is the list of numbers definable in
L, then Cantor'snumber for that list is not definable in L,
though it may be definablein some other language. It is
definable in any language in which thelist itself can be
defined. This creates no problem with Cantor'sargument in
general.===
===
Subject:
: Multiset TerminologyI'm currently doing
some research for my (computer science) PhD which islooking at
characteristics of agents in a system and the discovery
ofidentity based on these characteristics.As such, I've found
my way to multisets. I'm interested in the number ofrepeated
elements which a multiset has. For example, if you take the
setof agents in a system and a correlating set of their ages
then you couldhave n agents with n different age
possibilities. Alternatively you couldhave n agents with only
two or three possible values for age. Thisobviously affects
whether it is possible to identify an agent by
thecharacteristic in question.However, I'm not sure of the
terminology which I should use for searchingand working with
this property. Is there a term for this aspect of amultiset?
Can anyone point me to a reference or paper which could let
melook into this more deeply? Or even suggest a more standard
approachwhich would express this concept in a more useful
manner.Any information gratefully received.Joss
Wright===
===
Subject:
: Re: SCHOENFELDS RANDOM THEOREMTherefore the
probability that S does not occur in R is 0,> transistively,
the probability that S DOES occur in R is 1 - 0 = 1.PROOFWith
the greatest of respect, that is exactly what I thought your>
proof would end up demonstrating, and in fact by your use of
the word> 'random', it was the only sort of thing it _could_
end up> demonstrating.However if you choose an element at
random from an infinite set, the> probability of choosing any
particular element is zero. Hence proving> that an event has
probability zero does not prove that it is>
impossible.Undefined to be precise, since you must specify
drawing a numberbetween [n,m] from an infinite set bound by
[n,m]. Note, since it isan infinite set thenn = -INF AND/OR m
= +inf. If the probability of drawing this number from a
random location is 0,then the probability of not drawing it is
1. And therefore, you aresaying the probability of never
drawing it from randomly selectedlocations is always 1 (since
1^inf = 1).The problem here is the use of a non-axiomatic
definition ofprobability.P(x) = 1 states a certainty that
event x occurs.P(x) = 0 states a certainty that event x does
not occur. Without such axioms probability theory is a bit
useless.> Looks like the Nobel Prize for Mathematics has to be
postponed for yet> another year. However, transistively
speaking, you might get the Nobel> Prize for Electronics.>
Cheers,Zigoteau.===
===
Subject:
: Re: skewed dualities: functions
and sets>> |zzzzzzzzzzzzz ...>>> as i said, you probably
have to learn a fair amount of category theory>> to appreciate
this, so you're only demonstrating that you're an idiot.No, I'm
only demonstrating that I recognize a preposterous gasbag when
I >see one.Now is that fair, calling him a preposterous gasbag
just because he> used theorems from category theory to explain
why f^{-1} does not> commute with intersections?Hmm, maybe it
is.the last on my list of mathematical disciplines to
study?===
===
Subject:
: Re: skewed dualities: functions and sets>
|zzzzzzzzzzzzz ...as i said, you probably have to learn a fair
amount of category theory> to appreciate this, so you're only
demonstrating that you're an idiot.>>No, I'm only
demonstrating that I recognize a preposterous gasbag when I
>>see one.>>> Now is that fair, calling him a preposterous
gasbag just because he>> used theorems from category theory to
explain why f^{-1} does not>> commute with intersections?>>>
Hmm, maybe it is.>the last on my list of mathematical
disciplines to study?That depends on what sort of math you're
interested in.===
===
Subject:
: Re: skewed dualities: functions and
sets> R.Goldberg Methods of Real Analysis 2nd Edition
p11:f^-1(X union Y) = f^-1(X) union f^-1(Y)> f^-1(X intersect
Y) = f^-1(X) intersect f^-1(Y)> f(X union Y) = f(X) union
f(Y)However the relationf(X intersect Y) ?= f(X) intersect
f(Y)is <quote>Conspicuosly absent from this list</quote>is
some elementary perspective. I'm changing the notation,
because itis convenient to consider f: X-->Y, so that the
former X and Y setswould now become A and B, while X is domain
and Y image (as instandard notation).Let G be an arbitrary set
in cartesian product space XxY, and letpx(G) be projection,
i.e.x in px(G) iff there exists y such that (x,y) in GThen,
px(A / B) = px(A) / px(B)whilepx(A / B) != px(A) /
px(B)Therefore, union and intersection are not quite symmetric
as far ascartesian product is concerned. Next, function graph,
beingconstrained unsymmetrically in XxY along x and y somehow
skews theasymmetry even more...===
===
Subject:
: Re: Calculus 102: A
Humanistic Approach to Primitive Concepts by
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id i2FIUFP21841;Watashi wa Amerika-jin
desu. Nihon-go ga yoku wakarimasu-en, although as you can see I
cant spell it. :D===
===
Subject:
: Re: Size Frequency Distribution of
Large Impact Craters by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
i2FIglc23362;>log10(N) = a0 + lim(n = 1 to 11 )
an[log10(D)]^nthis is a mistake it should be:>log10(N) = a0 +
sigma(n = 1 to 11 ) an[log10(D)]^nalso when I say meteorite I
really mean crater diameter because different size meteorites
can make the same size crater.===
===
Subject:
: Re: discontinuities
of the limit of a sequence of functions>Suppose a sequence of
continuous functions (f_n) converges on a subset>D of R^n to a
real valued function f. If the convergence is uniform,>then f
is continuous, but if this is not the case, then is
there>anything interesting we can affirm about the set of
discontinuities of>f on D? I think this set has measure zero.
Is that true?No. For example, let S be a fat Cantor set: a
closed nowhere dense set> of positive measure. The set of
discontinuities of the indicator function > I_S(x) = 1 if x in
S, 0 otherwise, is S, and I_S can be obtained as the > limit of
a sequence of continuous functions> f_n(x) = 1 - n dist(x,S) if
dist(x,S) <= 1/n, 0 otherwise.Just to push this a little
further: the set of discontinuities can have full measure.
Choose a countalbe set {x1, x2, ...} = D, dense in R^n, and a
sequence of open sets Vm containing D with m(Vm) < 1/m. Let Em
= R^n Vm. The Em's are closed, so there are continuous fm : R^n
-> [0,1] with fm = 1 on E1 U ... U Em, f(xm) = 0, and fm > 0
elsewhere*. Set gm = f1*f2*...*fm. Then g1 >= g2 > = ..., so
the sequence gm converges pointwise everywhere to a function
g. We have g > 0 on U Em and g(xj) = 0 for each j, so g is
discontinuous at every point of U Em, a set of full
measure.*For example, let r be the distance from E1 U ... U Em
to xm; define a function h on [0,oo) by h(t) = t/r, 0<=t<=r,
h(t) = 1 for t > r; set fm(x) = h(|x-xm|).===
===
Subject:
: Re:
Question about associativity of cartesian product> For
example, are you really sure Z is a subset of Q and Q is a
subset of > R and R is a subset of C ?I would suggest you
never speak of that again. I brought up thatdistinction in
sci.logic a few weeks ago and people debated it sovigorously
that I ended up doing a sketch construction of the realsfrom Q
using equivalence classes of Cauchy sequences. Even after
Ihighlighted the difference between equivalence classes of
constantCauchy sequences of rational numbers and rational
numbers, they didn'tget it. :('cid 'ooh===
===
Subject:
: Re:
Question about associativity of cartesian product>As I
understand it (according
to>http://en.wikipedia.org/wiki/Cartesian_product),>1. the
cartesian product of two sets gives a set of 2-tuples
(ordered>pairs), i.e. X*Y={(x,y):x in X and y in Y}>2.
generally, the cartesian product of n sets gives a set of
n-tuples>3. cartesian product is associative. >But that third
statement seems to contradict the first two: by the>first two,
(A*B)*(C*D) should give a set of pairs of sets of pairs,>but by
the third statement, that same expression should give a set
of>4-tuples.Correct.>Obviously I'm missing something
obvious. Any enlightenment would be>appreciated.They are not
equal,To make sure I understand you correctly, please clarify
what yourantecedent for They is. Is it the cartesian products
(A*B)*(C*D)and (A*B*C*D)?> they are *isomorphic*, that is
there is a> *canonical, uniquely defined* isomorphism (that
is, a set bijection)I need to split hairs here; after further
study of this topic, I'mstill confused about isomorphisms, and
am unable to findcomprehensible (to me) answers to my
questions.1. Your parenthetical expression should be (that is,
a set bijectionwhich is a homomorphism whose inverse is also a
homomorphism), whichyou merely shortened for the sake of
brevity, correct?2. Assuming I have #1 right (according
tohttp://en.wikipedia.org/wiki/Isomorphism), this also raises
a sidequestion: is there such a thing as a set bijection which
is ahomomorphism whose inverse is NOT also a homomorphism? I'm
unable tothink of one. This raises a further side question: is
there such athing as a homomorphism whose inverse is also a
homomorphism, but isNOT a bijection? I'm unable to think of
one of these either.3. Please explain what you mean by
canonical, uniquely definedisomorphism here. I know what the
words mean, but do you mean thatthere's only one isomorphism
defined, or that the isomorphism which isdefined is the only
possible one?> between those two cartesian products.
Associativity is to be> understood in this sense, that
is(A*B)*C = A*(B*C)where = means there exists a bijection
(uniquely defined) between the> LHS and the RHS.well, does
uniquely defined mean that only one is defined or thatthe one
which is defined is the only possible one?2. Are the tuples
((a,b),(c,d)) and (a,(b,(c,d))) equal? I presumethat the
answer is no, and that a function such as reverse_tuple
wouldgive different results for them. When you write
Associativity is tobe understood in this sense (that = is
defined as isomorphic ratherthan equal), that seems to me to
violate the definition ofassociativity. For any function f and
any associative function
g,f(g(g(A,B),g(C,D)))=f(g(A,g(B,g(C,D)))). But this equality
is brokenif f is, for example, rst (a version of reverse which
reverses everytuple in a set), and g is cartesian product. It
seems that the problemcould be solved by defining cartesian
product to be normalizing,i.e. that it recursively computes
the cartesian product of any sets oftuples passed to it as
arguments, so that the final result is a set oftuples none of
whose (tuples is the antecedent here) elements is aset of
tuples, but of course then it would mangle original
argumentswhich happened to be sets of tuples for some other,
independentreason. And in any case, that's not how cartesian
product is defined.The problem seems to be that the
isomorphism in question is anisomorphism only with respect to
cartesian product. I'm having troubleformulating that last
sentence correctly, due to confusion about whatexactly
cartesian product is; it seems to be an external operation(as
that phrase is defined at the bottom
ofhttp://en.wikipedia.org/wiki/Binary_operation but with the
obviousgeneralization from a binary operation to an operation
of any arity),but that page then says An external binary
operation mayalternatively be viewed as an action with a link
from action tohttp://en.wikipedia.org/wiki/Group_action which
contains explanationsabout actions as they apply to groups
which confuse me thoroughly, andmy feeble brain is beginning
to want to jump out of my head.Further assistance would be
greatly appreciated.===
===
Subject:
: Re: Question about
associativity of cartesian product>[...]> I need to split
hairs here; after further study of this topic, I'm> still
confused about isomorphisms, and am unable to find>
comprehensible (to me) answers to my questions.> 1. Your
parenthetical expression should be (that is, a set bijection>
which is a homomorphism whose inverse is also a homomorphism),
which> you merely shortened for the sake of brevity, correct?>
2. Assuming I have #1 right (according to>
http://en.wikipedia.org/wiki/Isomorphism),While the examples
at wikipedia are o.k., their wording of thedefinition is a bit
misleading.You usually start with some notion of sets with
structure and acorresponding notion of structure preserving
maps which are calledhomomorphisms. The latter notion should
be defined in a such a waythat all identity maps are
homomorphisms and the composite of twohomomorphisms is again a
homomorphism.(this situation is in fact a special case of a
moregeneral situation idea, see
e.g.<http://en.wikipedia.org/wiki/Category_theory>)Now one
defines: a homomorphism f: X --> Y is an isomorphism for
whicha two sided inverse (w.r.t. composition) exists i.e., a
homomorphismg : Y ---> X such that f o g = id_Y and g o f =
id_X.If such a g exists, it follows that f is bijective and g
is indeedthe inverse map of f. The important point is, that on
requires thatg is indeed a homomorphism, not just any odd
map.Of course, with the above observation one can rephrase the
definitionof isomorphism asbijective homomorphism such that its
inverse map is again a homomorphism.All this can be applied to
the special case of sets: here we have_no_ structure and allow
_every_ set map as a homomorphism, simplybecause there is no
structure that needs to be preserved.Now the above definition
of isomorphism is equivalent to bijective map.In most
situations in algebra, the inverse map of an bijective
homomorphismwill already be a homomorphism (e.g. for groups,
rings or vector-spaces).This explains, why older algebra texts
often defined isomorphismas bijective homomorphism.> this also
raises a side> question: is there such a thing as a set
bijection which is a> homomorphism whose inverse is NOT also a
homomorphism?You make take topological spaces and continous
maps:Let X={0,1} with the discrete topology (all sets open)and
Y={0,1} with the indiscrete topology (only Y and the empty set
are open).Look at the continous map f: X ---> Y with f(0)=0
and f(1)=1.Another example is provided by ordered sets and
order preserving mapsLet X=P({1,2})={ {}, {1}, {2}, {1,2} }
ordered by inclusionand Y={0,1,2,3} ordered via 0<1<2<3.Look
at the order preserving map f: X ---> Y defined viaf({})=0,
f({x})=x (for x=1 or 2), and f({1,2})=3> This raises a further
side question: is there such a> thing as a homomorphism whose
inverse is also a homomorphism, but is> NOT a bijection? I'm
unable to think of one of these either.Your intuition is
correct this time.As mentioned above, any isomorphism has to
be bijective.> 3. Please explain what you mean by canonical,
uniquely defined> isomorphism here. I know what the words
mean, but do you mean that> there's only one isomorphism
defined, or that the isomorphism which is> defined is the only
possible one?One has to be careful here about the data that
describes theproduct of two (or more sets). In general on can
expect a lot ofdifferent isomorphism between (A*B)*C and
A*(B*C).For example, suppose |A|=|B|=|C|=2. Then both product
sets haveeight elements, so one can always compose any given
isomorphismf: (A*B)*C ---> A*(B*C) with some permutation of
A*(B*C) to obtainanother isomorphism.The crucial point is,
that product sets come together withprojection maps onto their
factors(just like those from the plane to the coordinate
axes).Given such projections p1: A*B ---> A and p2: A*B --->
Bin order to specify a map map f: X ---> A*B , it is enough
tospecify what the composites p1 o f and p2 o f should be
i.e.,to specify maps f1: X ---> A and f2: X ---> B.Now
suppose, we use the obvious projection mapsA <--- A*B ---> B (
a <- (a,b) -> b )B <--- B*C ---> C ( b <- (b,c) -> c )andA <---
A*(B*C) ---> B*C ( a <- (a,(b,c)) -> (b,c) )A*B <--- (A*B)*C
---> C ( (a,b) <- ((a,b),c) -> c )The canonical isomorphism
from (A*B)*C to A*(B*C) is then definedvia defining two maps
f1: (A*B)*C ---> A and f2: (A*B)*C ---> B*Cas follows:f1 is
built from the above projections via compositionf1 = (A*B)*C
---> A*B ---> A, f1( ((a,b),c) ) = ato specify f2 one can use
the same idea as above, namely specifyto maps f21: (A*B)*C
---> B and f22: (A*B)*C ---> C:f21 = (A*B)*C ---> A*B ---> B
f21( ((a,b),c) ) = bf22 = (A*B)*C ---> C f22( ((a,b),c) ) =
cThese combine to f2: (A*B)*C ---> B*C with f21( ((a,b),c) ) =
(b,c)Now f1 and f2 combine to f with f( ((a,b),c) ) =
(a,(b,c)).In a similar fashion you can produce a map g:
A*(B*C) ---> (A*B)*Cwhich is an inverse to f.This explains the
canonical --- but not yet the unique.For this, observe that
both A*(B*C) and (A*B)*C have certainprojections into the sets
A, B, and C, which are made from the aboveprojection maps:p1:
(A*B)*C ---> A , p1( ((a,b),c) ) = ap2: (A*B)*C ---> B , p2(
((a,b),c) ) = bp3: (A*B)*C ---> B , p3( ((a,b),c) ) = cq1:
A*(B*C) ---> A , p1( (a,(b,c)) ) = aq2: A*(B*C) ---> B , p2(
(a,(b,c)) ) = bq3: A*(B*C) ---> C , p3( (a,(b,c)) ) = cNow,
the above f satisfies qi o f = pi (i=1..3) and it isthe only
map that satisfies these equations.A similar result holds for
the g (to be constructed :-).Marc===
===
Subject:
: Re: Question
about associativity of cartesian product>>>As I understand
it (according
to>>http://en.wikipedia.org/wiki/Cartesian_product),>>1. the
cartesian product of two sets gives a set of 2-tuples
(ordered>>pairs), i.e. X*Y={(x,y):x in X and y in Y}>>2.
generally, the cartesian product of n sets gives a set of
n-tuples>>3. cartesian product is associative. >>But that
third statement seems to contradict the first two: by
the>>first two, (A*B)*(C*D) should give a set of pairs of sets
of pairs,>>but by the third statement, that same expression
should give a set of>>4-tuples.>>> Correct.>>>Obviously
I'm missing something obvious. Any enlightenment would
be>>appreciated.>>> They are not equal,>To make sure I
understand you correctly, please clarify what your>antecedent
for They is. Is it the cartesian products (A*B)*(C*D)>and
(A*B*C*D)?Yes.>> they are *isomorphic*, that is there is a>>
*canonical, uniquely defined* isomorphism (that is, a set
bijection)>I need to split hairs here; after further study of
this topic, I'm>still confused about isomorphisms, and am
unable to find>comprehensible (to me) answers to my
questions.>1. Your parenthetical expression should be (that
is, a set bijection>which is a homomorphism whose inverse is
also a homomorphism), which>you merely shortened for the sake
of brevity, correct?Sorry, I'm using the word isomorphism in
the spirit of categorytheory. Isomorphism here just means
isomorphism in the category ofsets which is just a bijection
of sets -- no algebraic structure orother is being implied.>2.
Assuming I have #1 right (according
to>http://en.wikipedia.org/wiki/Isomorphism), this also raises
a side>question: is there such a thing as a set bijection which
is a>homomorphism whose inverse is NOT also a homomorphism? I'm
unable to>think of one. This raises a further side question: is
there such a>thing as a homomorphism whose inverse is also a
homomorphism, but is>NOT a bijection? I'm unable to think of
one of these either.>3. Please explain what you mean by
canonical, uniquely defined>isomorphism here. I know what the
words mean, but do you mean that>there's only one isomorphism
defined, or that the isomorphism which is>defined is the only
possible one?What I mean by canonical is that it commutes with
the projections.In a cartesian product there are
projectionsp_A: A*B -> A defined by (a,b) |-> ap_B:A*B -> B
defined by (a,b) |-> bLikewise for cartesian products (A*B)*C,
A*(B*C) there are projectionsinto each of its factors. The
canonical isomorphismi:(A*B)*C -> A*(B*C)commutes with the
projections. In details: we have the equality offunctionsp_A =
p_A compose iwhere p_A are the relevant projections of the
cartesian products, andthe similar equations for p_B and p_C.
In fact i is the *unique* map(A*B)*C -> A*(B*C) that commutes
with the projections - that is what Imean by being uniquely
defined. It is a theorem by S. Maclane thatgiven two ways to
parenthesize an n-fold cartesian product (forexample, for n=3,
there are only 2 ways: (A*B)*C and A*(B*C) but forn=4 there are
already 5 ways) there is a *unique* isomorphism betweenthem
commuting with the projections. This justifies the usual
practiceof treating the cartesian product of sets as if it
were a strictlyassociative operation.WIth my best regards,G.
Rodrigues===
===
Subject:
: Re: Question about associativity of
cartesian product>It is a theorem by S. Maclane that>given two
ways to parenthesize an n-fold cartesian product (for>example,
for n=3, there are only 2 ways: (A*B)*C and A*(B*C) but
for>n=4 there are already 5 ways) there is a *unique*
isomorphism between>them commuting with the
projections.Disregard this. In this context, and in the way
I've stated it,Maclane's theorem is a triviality. In the wider
context (of monoidalcategories) Maclane's theorem is both
non-trivial and useful. And inthe way I've stated neither it
makes sense (there are no projectionsin monoidal categories to
speak of) nor it is correct (it is *not*true that the canonical
isomorphism is unique - it's a bit moresubtle).I'm sure there
was a point in introducing Maclane's theorem at thisstage,
except I forgot which was it. With my best regards,G.
Rodrigues===
===
Subject:
: Re: Question about associativity of
cartesian product Adjunct Assistant Professor at the
University of Montana.Pet peeve:>Disregard this. In this
context, and in the way I've stated it,>Maclane's theorem is a
triviality. In the wider context (of monoidal>categories)
Maclane's theorem is both non-trivial and useful.The proper
spelling is Mac Lane (with a space between the two).--
==============================================================
========It's not denial. I'm just very selective about what I
accept as reality. --- Calvin (Calvin and
Hobbes)=======================================================
==================
===
Subject:
: Re: Question about associativity of
cartesian product>>As I understand it (according
to>>http://en.wikipedia.org/wiki/Cartesian_product),>>1. the
cartesian product of two sets gives a set of 2-tuples
(ordered>>pairs), i.e. X*Y={(x,y):x in X and y in Y}>>2.
generally, the cartesian product of n sets gives a set of
n-tuples>>3. cartesian product is associative. >>But that
third statement seems to contradict the first two: by
the>>first two, (A*B)*(C*D) should give a set of pairs of sets
of pairs,>>but by the third statement, that same expression
should give a set of>>4-tuples.>>> Correct.>>>Obviously
I'm missing something obvious. Any enlightenment would
be>>appreciated.>>> They are not equal,> To make sure I
understand you correctly, please clarify what your> antecedent
for They is. Is it the cartesian products (A*B)*(C*D)> and
(A*B*C*D)?Right>> they are *isomorphic*, that is there is a>>
*canonical, uniquely defined* isomorphism (that is, a set
bijection)> I need to split hairs here; after further study of
this topic, I'm> still confused about isomorphisms, and am
unable to find> comprehensible (to me) answers to my
questions.> 1. Your parenthetical expression should be (that
is, a set bijection> which is a homomorphism whose inverse is
also a homomorphism), which> you merely shortened for the sake
of brevity, correct?No. It is just a set bijection and nothing
else. You cannot speakabout homomorphisms, since there is no
algebraic structure here.> 2. Assuming I have #1 right
(according to> http://en.wikipedia.org/wiki/Isomorphism), this
also raises a sideThere's a big confusion here. As it is
written at that web page,isomorphisms are structure-preserving
mappings. But there's nostructure to preserve here. Up until
now, you were just speakingabout sets.> question: is there
such a thing as a set bijection which is a> homomorphism whose
inverse is NOT also a homomorphism? I'm unable to> think of
one. This raises a further side question: is there such a>
thing as a homomorphism whose inverse is also a homomorphism,
but is> NOT a bijection? I'm unable to think of one of these
either.Tell me first what algebraic structure you are talking
about.> 3. Please explain what you mean by canonical, uniquely
defined> isomorphism here. I know what the words mean, but do
you mean that> there's only one isomorphism defined, or that
the isomorphism which is> defined is the only possible one?The
second one, of course, since no isomorphism was defined here.> between those two cartesian products. Associativity is to
be>> understood in this sense, that is>>> (A*B)*C =
A*(B*C)>>> where = means there exists a bijection (uniquely
defined) between the>> LHS and the RHS.Yes> well, does
uniquely defined mean that only one is defined or that> the
one which is defined is the only possible one?Again, no
isomorphism was actually defined. That excludes the
firstpossibility.> 2. Are the tuples ((a,b),(c,d)) and
(a,(b,(c,d))) equal?No.Best regards,Jose Carlos
Santos===
===
Subject:
: Re: This Week's Finds in Mathematical
Physics (Week 202)X-MailScanner-SpamxCheck: not spam,
SpamAssassin (score=-6.6, required 5, REFERENCES)>Here's a
neat 18th century way of getting the Cat numbers.>Consider the
equation for Lisp S-expressions>S = A + S x S>meaning that an
S-expression is either an atom or a pair of>S-expressions.
Regard this as a quadratic equation for S and solve it>by the
quadratic formula. Expand the square root in power series.>The
coefficients of the powers of A are just the Cat numbers>giving
the number of types of S-expression with a given number>of
atoms. Of course, you get set isomorphisms as well.>I suppose
this is a special case of John Baez's theory.Yes! But, this
part of the theory isn't
mine..........................................................
...............Also available at
http://math.ucr.edu/home/baez/week144.html[...]Suppose we want
to know how many ways there are to multiplyn things in a
nonassociative algebra, not counting differentways of
permuting these things. So, we want to bracketings of n
letters that correspond to binary planar trees with n leaves.
For example, when n = 4: a b c d / / / / / / / / ((ab)c)d / /
/ / a b c d / / / / / / (a(bc))d / / / / a b c d / / / / /
(ab)(cd) / / / a b c d / / / / / a((bc)d) / / / a b c d / / /
a(b(cd)) / / /We can think of these trees as recording the
*process* of multiplying n things, with time marching down the
page. How many binary planar trees with n leaves are there,
anyway? Well,the answer is called the (n-1)st Cat number.
These numberswere first discovered by Euler, but they're named
after Eugene Cat, who discovered their relation to binary
trees. Here they are, starting from the 0th one:1, 1, 2, 5,
14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900,
...The nth Cat number is also the number of ways of taking a
regular(n+2)-gon and chopping it into triangles by connecting
the vertices byline segments that don't cross each other. It's
also the number of waysof getting from a street corner in
Manhattan to another street cornerthat's n blocks north and n
blocks east, always driving north or east, but making sure
that at no stage have you gone a greater total distance north
than east. Get it? No? Maybe a picture will help! When n = 3,
there are 5 ways: . . . . . . . . . . . . . . . . . . . . | |
| | | . . . . . . . . . . .__. . . . . . . .__. | | | | | . .
. . . . .__. . . . . . .__.__. . .__. . | | | | | .__.__.__.
.__.__. . .__.__. . .__. . . .__. . .I leave it as a puzzle
for you to understand why all these things arecounted by the
Cat numbers. If you want to see nicer pictures ofall these
things, go here:7) Robert M. Dickau, Cat numbers,
http://forum.swarthmore.edu/advanced/robertd/cat.htmlFor more
problems whose answer involves the Cat numbers, try this:8)
Kevin Brown, The meanings of Cat numbers,
http://www.seanet.com/~ksbrown/kmath322.htmTo figure out a
formula for the Cat numbers, we can use thetechnique of
generating functions:9) Herbert Wilf, Generatingfunctionology,
Academic Press, Boston, 1994. Also available free at
http://www.cis.upenn.edu/~wilf/Briefly, the idea is to make up
a power series T(x) where the coefficientof x^n is the number
of n-leaved binary trees. Since by some irritatingaccident of
history people call this the (n-1)st Cat number, wehave:T(x) =
sum C_{n-1} x^nWe can do this trick whenever we're counting how
many structures of some sort we can put on an n-element set.
Nice operations on structures correspond to nice operations on
formal power series. Using thiscorrespondence we can figure out
the function T and then do a Taylor expansion to determine the
Cat numbers. Instead of explaining the theory of how this all
works, I'll just demonstrate it as a kind of magic trick.So:
what is a binary tree? It's either a binary tree with one leaf
(the degenerate case) or a pair of binary trees stuck together.
Now let's translate this fact into an equation:T = x + T^2Huh?
Well, in this game plus corresponds to or, times corresponds
to and, and the power series x is the generating function for
binarytrees with one leaf. So this equation really just says a
binary tree equals a binary tree with one leaf or a binary tree
and a binary tree.Next, let's solve this equation for T. It's
just a quadratic equation,so any high school student can solve
it:T = (1 - sqrt(1 - 4x))/2.Now if we do a Taylor expansion we
getT = x + x^2 + 2x^3 + 5x^4 + 14x^5 + 42x^6 + ...Lo and
behold - the Cat numbers! If we're a bit smarter and use the
binomial theorem and mess around a bit, we get a closed-form
formula for the Cat numbers:C_n = (2n choose n)/(n + 1)Neat,
huh? If you want to understand the category-theoretic
foundations of this trick, read about Joyal's concept of
species.This makes precise the notion of a structure you can
put on a finite set. For more details, try:10) Andre Joyal,
Une theorie combinatoire des series formelles, Adv. Math. 42
(1981), 1-82.11) F. Bergeron, G. Labelle, and P. Leroux,
Combinatorial species and tree-like structures, Cambridge,
Cambridge U. Press, 1998.Footnote - If you want to know more
about the deep inner meaningof the Cat numbers, try this:12)
Richard P. Stanley, Enumerative Combinatorics, volume
2,Cambridge U. Press, Cambridge, 1999, pp. 219-229.It lists
*66* different combinatorial interpretations of thesenumbers!
As an exercise, it urges you to prove that they allwork,
ideally by finding 4290 simple and elegant bijections between
the various sets being counted. ===
===
Subject:
: Searching for
Ptolemy Math CardsHi+ I recently found a buisness card sized
card in the glove compartmentof a recent rental car in
California. On one side it depicted astrange tesselating image
and on the other it said the following:[begin quote]Strauss
333The pattern is one of only 17 possible planar symetries
made byreplicating a finite motif; this particular symmetry is
completelydetermined by the three kinds of points about which
the pattern can berotated by 120 degrees: the centers of the
heads, just off the chinsand between the noses. The notation
333 refers, then, to the threekinds of 3-fold centers of
rotation.What would happen if you folded the pattern up,
somehow, so that allthe noses coincide, all the eye-brows
match up, etc? For starts you'dget a fat wad of paper! SO
suppose the pattern is on an infinitelythin sheet; then after
folding you'd have a tidy little surface withone nose, one
eye, etc.This resulting surface is called an orbifold. THe
orbifold for the333 symmetry turns out to be a triangular
pillow shape. Suchorbifolds provide a nice way to classify
symmetries.[end quote]It also says serving kids since 1953 and
Did yo know? The Ptolemymathcard co. is a wholly owned
subsidiary of the Cadian Borax Corp!Well, I thought this was
pretty cool, so I kept it, but the strangething is I've been
unable to find any mention of a company with thisname. To make
matters stranger, when a friend of mine was travellingin Madrid
he found a Ptolemy Mathcard inserted into the Bible in hishotel
room! His had a photograph of many other geometrically
arrangedmathcards (including mine and itself) on one side and
on the other itsaid:[begin quote]PolyptolemyThe cards lie on
the edges of a regular icosahedron as well as theedges of a
regular dodecahedron. Cards that are alike lie on theedges of
a regular tetrahedron, the faces of a cube and the verticesof
a regular octahedron. Altogether, this reveals the cosets of
thetetrahedral group in the icosahedral group.In another vein,
consider all the points in the scene that lie in aninfinite
sequence of nested Polyptolemy cards. These points form
aCantor set, a most remarkable entity. Cantor sets are
completelydisconnected, but uncountable sets.Amazingly the
Canotro set on the surface of the card hasinaccessible points;
that is even though the set seems to bevanishingly dusty, there
are points that can be isolated from theoutside of the Cantor
set.The rest is art.[end quote]However this card said serving
kids since 2248 and Did you know? Ptolemy mathcards are
excellent for unclogging drains!I'm wondering if anyone else
has found any of these peculiar cards.===
===
Subject:
: Re:
Searching for Ptolemy Math CardsNNTP-Posting-User:
[rXawWTU8qP1BOMgpUIEuefwqm2vDec9Z]> Hi+> I recently found a
buisness card sized card in the glove compartment> of a recent
rental car in California. On one side it depicted a> strange
tesselating image and on the other it said the following:>
[begin quote]> Strauss 333> The pattern is one of only 17
possible planar symetries made by> replicating a finite motif;
this particular symmetry is completely> determined by the three
kinds of points about which the pattern can be> rotated by 120
degrees: the centers of the heads, just off the chins> and
between the noses. The notation 333 refers, then, to the
three> kinds of 3-fold centers of rotation.> What would happen
if you folded the pattern up, somehow, so that all> the noses
coincide, all the eye-brows match up, etc? For starts you'd>
get a fat wad of paper! SO suppose the pattern is on an
infinitely> thin sheet; then after folding you'd have a tidy
little surface with> one nose, one eye, etc.> This resulting
surface is called an orbifold. THe orbifold for the> 333
symmetry turns out to be a triangular pillow shape. Such>
orbifolds provide a nice way to classify symmetries.> [end
quote]> It also says serving kids since 1953 and Did yo know?
The Ptolemy> mathcard co. is a wholly owned subsidiary of the
Cadian Borax Corp!Have you tried a Google search for ptolemy
mathcard?-- J. H. PalmieriDept of Mathematics, Box 354350
mailto:palmieri@math.washington.eduUniversity of Washington
http://www.math.washington.edu/~palmieri/Seattle, WA
98195-4350===
===
Subject:
: Math Day ideasI am trying to organize a
math day in our campus, Indiana University Kokomo (Kokomo,
Indiana). I want to invite Juniors and Seniors from local
schools. I want to organize an entertaining, educational talk
as part of the activities. If you have done some thing like
this on your campus, will you please share your experiences?
If you have any ideas about this please let post or e-mail to
me. If you know of any one who can speak to these high school
students about mathematics, physics, or chemistry and their
marvelous applications, please let me know how I can contact
him. Your help in this matter is greatly appreciated. Thank
you very much. .. Raghu Gompa (rgompa@iuk.edu)===
===
Subject:
:
Range of Entire function charset=Windows-1252By Picard's
Little Theorem, the range of the function exp(z) is C-{0},
sothe inverse image of any point w, is either an infinite set
(for any w=/=0)or the null set (for 0). (naturally)What about
the function f(z)=z*exp(z)? exp(z) misses at most {0},
but0*exp(0)=0, so the range of f doesn't miss any
points.Picard's Little Theorem _for f_ (since f is entire
also) implies that theinverse image of any w under f is either
an infinite set or the null setalso.What are the rest of values
(infinitely many of them) in the inverse imageof {0} under
f?What am I missing here? I only see ONE point being the
inverse image of {0}and that's neither the null set nor an
infinite set.--Ioannis
Galidakishttp://users.forthnet.gr/ath/jgal/-------------------
-----------------------Eventually, _everything_ is
understandable===
===
Subject:
: Re: Range of Entire function>By
Picard's Little Theorem, the range of the function exp(z) is
C-{0}, so>the inverse image of any point w, is either an
infinite set (for any w=/=0)>or the null set (for 0).
(naturally)I'd say that's not a result of Picard's Little
Theorem, but it is an illustration of it.>What about the
function f(z)=z*exp(z)? exp(z) misses at most {0},
but>0*exp(0)=0, so the range of f doesn't miss any points.Not
necessarily. There's no relation between the points f_1(z) and
f_2(z)miss and the points f_1(z) f_2(z) misses.>Picard's Little
Theorem _for f_ (since f is entire also) implies that
the>inverse image of any w under f is either an infinite set
or the null set>also.No, that's not at all what it says. And z
exp(z) is the perfect counterexample to your statement.>What
are the rest of values (infinitely many of them) in the
inverse image>of {0} under f?>What am I missing here? I only
see ONE point being the inverse image of {0}>and that's
neither the null set nor an infinite set.Exactly true.
Picard's theorem says the inverse image is infinite with at
most one exception.It doesn't say that the inverse image for
that exception can't be finiteand nonempty.Department of
Mathematics http://www.math.ubc.ca/~israel University of
British Columbia Vancouver, BC, Canada V6T 1Z2===
===
Subject:
: Re:
Range of Entire function charset=iso-8859-7 Robert Israel
<israel@math.ubc.ca>   >What about the function f(z)=z*exp(z)?
exp(z) misses at most {0}, but>0*exp(0)=0, so the range of f
doesn't miss any points.> Not necessarily. There's no relation
between the points f_1(z) and f_2(z)> miss and the points
f_1(z) f_2(z) misses.You are of course right in general, but
in this case I think the range IS C.See my reply to
David.>What am I missing here? I only see ONE point being the
inverse image of{0}>and that's neither the null set nor an
infinite set.> Exactly true.> Picard's theorem says the
inverse image is infinite with at most one> exception.> It
doesn't say that the inverse image for that exception can't be
finite> and nonempty.Doh! Thank you Robert. That's what I was
looking for. Guess it's time forsome more sleep :*)>
Department of Mathematics http://www.math.ubc.ca/~israel>
University of British Columbia> Vancouver, BC, Canada V6T
1Z2--Ioannis
Galidakishttp://users.forthnet.gr/ath/jgal/-------------------
-----------------------Eventually, _everything_ is
understandable===
===
Subject:
: Re: Range of Entire function>By
Picard's Little Theorem, the range of the function exp(z) is
C-{0}, so>the inverse image of any point w, is either an
infinite set (for any w=/=0)>or the null set (for 0).
(naturally)>What about the function f(z)=z*exp(z)? exp(z)
misses at most {0}, but>0*exp(0)=0, so the range of f doesn't
miss any points.How do you know the range of f is all of C?
I'm not saying it'snot, I just don't see why it's obviously
true.>Picard's Little Theorem _for f_ (since f is entire also)
implies that the>inverse image of any w under f is either an
infinite set or the null set>also.??? How does this follow
from Picard's Little Theorem?>What are the rest of values
(infinitely many of them) in the inverse image>of {0} under
f?>What am I missing here? Hard to say, since you're leaving
out the proofs. You seem to besaying that PLT shows that for
any (non-polynomial) entire f andany w, the inverse image of w
under f is either empty or infinite.How does PLT imply that?>I
only see ONE point being the inverse image of {0}>and that's
neither the null set nor an infinite set.===
===
Subject:
: Re: Range
of Entire function charset=utf-8 <ullrich@math.okstate.edu>
[CapitalEth][EDoubleDot][Micro] .b3 .b9[EDoubleDot].b9>What
about the function f(z)=z*exp(z)? exp(z) misses at most {0},
but>0*exp(0)=0, so the range of f doesn't miss any points.>
How do you know the range of f is all of C? I'm not saying
it's> not, I just don't see why it's obviously true.The
inverse of z*exp(z) is the multivalued LambertW function
(denoted byLW), which is defined everywhere in C and
satisfies:LW(z)*exp(LW(z))=z.So, given w in C, LW(w) [in fact
there are infinitely many values, since LWis multivalued] is a
pre-image of w under f(z)=z*exp(z).(Then again, I know you are
allergic to this function. Don't start asking mehow it's
defined and things...)>Picard's Little Theorem _for f_ (since
f is entire also) implies that the>inverse image of any w
under f is either an infinite set or the null set>also.> ???
How does this follow from Picard's Little
Theorem?http://www.math.niu.edu/~rusin/known-math/95/picard>
What are the rest of values (infinitely many of them) in the
inverseimage>of {0} under f?What am I missing here?> --Ioannis
Galidakishttp://users.forthnet.gr/ath/jgal/--------------------
----------------------Eventually, _everything_ is
understandable===
===
Subject:
: Re: Range of Entire function>
<ullrich@math.okstate.edu> [CapitalEth][EDoubleDot][Micro] .b3
.b9[EDoubleDot].b9>Picard's Little Theorem _for f_ (since f
is entire also) implies that the>inverse image of any w under
f is either an infinite set or the null set>also.>>??? How
does this follow from Picard's Little Theorem?>http://www.math.niu.edu/~rusin/known-math/95/picard> The
result as stated there, viz.,Little Picard Theorem: Every
complex analytic function defined on the whole>complex plane
is either> (a) onto C (i.e., for every w there is a z with
f(z)=w)> (b) onto C-{one point} (i.e., f misses at most one
value)> (c) or constant.>In fact, in all cases, the inverse
image of every point is either empty or>infinite.Is clearly
false. Take, for instance, f(z) = z, contradicting the last
sentence. Care to make a correction, Dr. Rusin?-- Stephen J.
Herschkorn herschko@rutcor.rutgers.edu===
===
Subject:
: Re: Range of
Entire function charset=utf-8 Stephen J. Herschkorn
<herschko@rutcor.rutgers.edu> [CapitalEth][EDoubleDot][Micro]
.b3
.b9[EDoubleDot].b9>http://www.math.niu.edu/~rusin/known-math/
95/picard> The result as stated there, viz.,>Little Picard
Theorem: Every complex analytic function defined on
thewhole>complex plane is either> (a) onto C (i.e., for every
w there is a z with f(z)=w)> (b) onto C-{one point} (i.e., f
misses at most one value)> (c) or constant.>In fact, in all
cases, the inverse image of every point is either
emptyor>infinite.> Is clearly false. Take, for instance, f(z)
= z, contradicting the last> sentence. Care to make a
correction, Dr. Rusin?I think that Dave means: Every
_non-polynomial_ complex analytic functiondefined... etc.> --
> Stephen J. Herschkorn herschko@rutcor.rutgers.edu--Ioannis
Galidakishttp://users.forthnet.gr/ath/jgal/-------------------
-----------------------Eventually, _everything_ is
understandable===
===
Subject:
: Re: Range of Entire function>
Stephen J. Herschkorn <herschko@rutcor.rutgers.edu>
[CapitalEth][EDoubleDot][Micro]>.b3
.b9[EDoubleDot].b9>>http://www.math.niu.edu/~rusin/known-math/
95/picard>> The result as stated there, viz.,>>Little Picard
Theorem: Every complex analytic function defined on
the>whole>>complex plane is either>> (a) onto C (i.e., for
every w there is a z with f(z)=w)>> (b) onto C-{one point}
(i.e., f misses at most one value)>> (c) or constant.>>In
fact, in all cases, the inverse image of every point is either
empty>or>>infinite.>> Is clearly false. Take, for instance,
f(z) = z, contradicting the last>> sentence. Care to make a
correction, Dr. Rusin?Did you read the whole page?>I think
that Dave means: Every _non-polynomial_ complex analytic
function>defined... etc.But in fact if you continue reading,
he does correct himself in thatvery page, and state that the
In fact, ... is incorrect, even fornon-polynomials. Department
of Mathematics http://www.math.ubc.ca/~israelUniversity of
British ColumbiaVancouver, BC, Canada V6T 1Z2===
===
Subject:
: Re:
Range of Entire function>By Picard's Little Theorem, the range
of the function exp(z) is C-{0}, so>the inverse image of any
point w, is either an infinite set (for any w=/=0)>or the null
set (for 0). (naturally)>What about the function f(z)=z*exp(z)?
exp(z) misses at most {0}, but>0*exp(0)=0, so the range of f
doesn't miss any points.>Picard's Little Theorem _for f_
(since f is entire also) implies that the>inverse image of any
w under f is either an infinite set or the null set>also.>What
are the rest of values (infinitely many of them) in the
inverse image>of {0} under f?>What am I missing here? I only
see ONE point being the inverse image of {0}>and that's
neither the null set nor an infinite set.According to Conway's
complex analysis text, the following is a corollary of the
Great Picard Theorem:If f is an entire function that is not a
polynomial, then f assumes every complex number, with one
exception, an infinite number of times.(BTW, this is the last
result in the text proper.) The proof considers g(z) = f(1/z),
which has an essential singularity at 0. Indeed, GPT assures
that g assumes each complex number, with one possible
exception, an infinite number of times. Thus, this exception
may appear any finite number of times.I don't remember much
complex analysis: Where did you get that the inverse image
must be empty if not infinite? Little Picard just says that
the range of a nonconstant entire function contains every
complex number with one possible exception. Foir example,
letting f(z) = z, f^(-1)(w) is a singleton for each value.--
Stephen J. Herschkorn herschko@rutcor.rutgers.edu===
===
Subject:
:
Re: Range of Entire function> If f is an entire function that
is not a polynomial, then f assumes> every complex number,
with one exception, an infinite number of times.Why every
function that's not a polynomial? You can write lots of
functionsin terms of an infinite Taylor series, which is like
an infinite degreepolynomial, why should this not as a
polynomial in that sense?===
===
Subject:
: Re: Range of Entire
function>>If f is an entire function that is not a
polynomial, then f assumes>>every complex number, with one
exception, an infinite number of times.>>Why every function
that's not a polynomial? You can write lots of functions>in
terms of an infinite Taylor series, which is like an infinite
degree>polynomial, why should this not as a polynomial in that
sense?Because if f is a polynomial of degree n, then, for any
w, there are at most n distinct values z such that f(z) = w.--
Stephen J. Herschkorn herschko@rutcor.rutgers.edu===
===
Subject:
:
Sector angle from segment areaHi there,Can anyone suggest a
formula for deriving sector angle from the areaof segment (of
that sector)?Thank you.Muxa.===
===
Subject:
: Re: Sector angle from
segment area> Can anyone suggest a formula for deriving sector
angle from the area> of segment (of that sector)?One other bit
of information (such as the radius r of the circle) must
begiven.In any event, please look
at<http://mathforum.org/dr.math/faq/faq.circle.segment.html>
first.Then, if you still have questions, ask here again.David
Cantrell===
===
Subject:
: Can someone help me with an algebra
problem?Hi all, this one's got me stuck, and I'm not sure if
it's because it'sparticularly hard, or because I'm just
braindead this week. Cansomeone just give me a
hint?Problem:Let R<S be Noetherian rings (comm. with ident.)
such that S isintegral over R. If R is local, prove that S has
only finitely manymaximal ideals.The first thing I notice is
that all maximal ideals of S must lie overthe maximal ideal of
R (Lang's Algebra, Prop VII.1.11), but I can'tget anywhere past
that. The only ascending chain I can think of wouldbe to take
the infinite intersection of all max ideals of S, then
allexcept one, all except two, etc., but that seems totally
useless anddoesn't use the maximal ideal of R anyway.I'm only
an amateur hack (or hobbyist), not a student, so theinternet
is my only hope. TIA-Rob===
===
Subject:
: Re: Rolling a Dice for
1000 times.> The probability of rolling a die and having it
come up 6> 100 times in a row is 1/6^100, or about 1.53 *
10^-78.That's about the reciprocal of the number of atoms in
the> Universe; if each of them were a die, and each of them
were> rolled 100 times simultaneously, then one of them --
somewhere --> might have shown 100 6's in a row. (The actual
probability> is not certainty, as it turns out. It's about
1/e.)So if you do have a die that rolls 100 times in a row
and> comes up 6, chances are it's loaded.How does that follow?
The probability of 100 1's is also 1/6^100, sois the
probability of 50 consecutive 4's followed by 50
consecutive3's, or throwing 123456 ... 1234, or any other
combination you canimagine. Why do some of these combinations
imply the dice is loadedwhen other's don't?Just
wondering,-Sunny===
===
Subject:
: Re: Rolling a Dice for 1000
times.> The probability of rolling a die and having it come up
6> 100 times in a row is 1/6^100, or about 1.53 * 10^-78.That's
about the reciprocal of the number of atoms in the> Universe;
if each of them were a die, and each of them were> rolled 100
times simultaneously, then one of them -- somewhere --> might
have shown 100 6's in a row. (The actual probability> is not
certainty, as it turns out. It's about 1/e.)So if you do have
a die that rolls 100 times in a row and> comes up 6, chances
are it's loaded.How does that follow? The probability of 100
1's is also 1/6^100, so> is the probability of 50 consecutive
4's followed by 50 consecutive> 3's, or throwing 123456 ...
1234, or any other combination you can> imagine. Why do some
of these combinations imply the dice is loaded> when other's
don't?Just wondering,How do you load a die in order for it to
roll a 1, then a 2, then a 3, then a 4, then a 5, then a 6? Of
course the is the dice loaded criterion should be a real-world
one -- if you can make money from a bookie by rolling the die,
bettingon its outcome, then it's loaded.-- 1st bug in MS win2k
source code found after 20 minutes: scanline.cpp2nd and 3rd
bug found after 10 more minutes: gethost.cBoth
non-exploitable. (The 2nd/3rd ones might be, depending on the
CRTL)===
===
Subject:
: Vector calculus and parametric equationsAn
object is at P[t] = {9 Sin[t], 3 Cos[t]}at time t.a. For 0 <=
t <= 2pi/3 the object is constrained to move on an ellipse.
Suppose, at the instant t = 2pi/3 the object is released from
all constraints and allowed to move of its own free will
independent of any accelerations due to forces. Plot the path
the object takes. (My guess is it goes off on a tangent line
from the ellipse. Am I correct?)b. The speed of the object at
time t is defined to be the length of the velocity vector at
time t; in other words speed[t] =
sqrt(velocity[t].velocity[t])Put unittan[t] =
tan[t]/sqrt(tan[t].tan[t])where tan[t] =
{x'[t],y'[t]}Calculate D[speed[t],t] (derivative of speed[t]
with respect to t) and compare it with
acceleration[t].unittan[t].Try to explain why you think that
the result is natural or weird. (The answers came out that
both are equal, but I really don't see what they're getting
at. What is unittan? And what is acceleration[t].unittan[t]?c.
If you know speed[0], then how do you give a formula for
speed[t] in terms of acceleration.unittan[t]?d. Can you get
any information about speed[t] if you know the tangential
component of acceleration but lose all your information about
the normal component of acceleration?===
===
Subject:
: Re: Cantor's
Diagonal Argument> So if one believes that proving this
theorem about some countable list> C of numbers and some
number b suggests that b is in the list C, one> has to think
that the square root of two is rational (indeed one hasBoth
issues deal with countability, but all_reals is about
existence of thenumber, rationality is about properties of the
number.We just use difference classes of 'belongs to the
list'.Rationals have other properties like repeating digits,
pi can be shown notto repeat, all rationals will demonstrate
this property.Herc===
===
Subject:
: Re: Cantor's Diagonal Argument>
So if one believes that proving this theorem about some
countable list> C of numbers and some number b suggests that b
is in the list C, one> has to think that the square root of two
is rational (indeed one has> Both issues deal with
countability, but all_reals is about existence of the> number,
rationality is about properties of the number.> We just use
difference classes of 'belongs to the list'.<unsnip for all n,
exists j, for all i> b(n)_i = c(j)_i where> b = DIAG> c =
COMPUTABLES> That's right, and they're different.agreed, but
is it significant?they both mean the number is on the list to
any degree required.the 1st digit is on the list,and the 2nd
digit is on the list too,and the 3rd digit is on the list
too....all digits are on the listtherefore the number is on
the list.</unsnip>The computables here are definately shown
not to be missing any numbers inthe ordinary
meaning.Any_number_of_matching_digits_can_be_found does not
imply to me thatthere are an infinitude higher of missing
numbers! does that imply that toyou? of course not. its a
simple self referencing representation issue, not asuper
infinity of missing numbers, super dense number classes, just
usecomputers to count them, that simple.The diag is defined
like this :Make the number not equal to the 1st numberand make
the number not equal to the second numberand make the number
not equal to the third number...That's its rather obviously
cyclic definition, that's not its propertiesbecause it's just
got a silly definition of not being a number at
all.Herc===
===
Subject:
: Re: Cantor's Diagonal Argument>> Ouch!!
Doesn't this mean b = c(j)? and is it correctly derived?>> for
all n, exists j, b_1 = c(j)_1, b_2 = c(j)_2, ... b_n = c(j)_n>>
This is equivalent to : for all n, b_n = c(j)_n>> You have to
prove that there is such a j.> note the exists j> for all
n, exists j, b_1 = c(j)_1, b_2 = c(j)_2, ... b_n = c(j)_n>
admitedly this is different to :> exists j, for all n, b_n =
c(j)_n> I showed that no matter how many digits you examine,
the diag> up to that many digits is covered on the list.>
HercYou have to show that for all n, there exists a j, such
that for all i,> b_n_i = c(j)_i>b_n is a particular b? I'll
write b(n) and keep _i notation for digit indexing.I have :for
all n, for all i, exists j b(n)_i = c(j)_iYou want :for all n,
exists j, for all i b(n)_i = c(j)_iwhereb = DIAGc =
COMPUTABLES> Yes, that's what he would need to show.> It might
be helpful though to express what he has shown in terms of>
what sort of nunmbers are involved, because it's often easier
to see> things expressed in simple categories instead of in
something like> predicate calculus. And to show how the same
thing can be proved> about simpler lists and rather more
obvious non-members.> So what does for all n, exists j, b_i =
c)j)_i for i in 1....n mean.> It means If x is the number
represented by b_1.....b_n, then some> number whose expression
has those n digits as the first n is on the> list. If the list
includes all numbers whose expansion terminates> after a
finite number of terms this is trivially true, since x itself>
is on the list. Now a list of all rational numbers certainly
includes> all numbers with finite expansions, so it's clear
that if the list is> the list of all rational numbers and b
(represented as b_1....) is any> real number then for any n we
can find the number b_1....b_n on the> list.> Think about the
well known fact, established by the ancient Greeks,> that
there is no rational number whose square is 2. Mow let C(j)
be> the complete set of rationals between 0 and 1 indexed by
natural> numbers j, and let b be the real number such that
(b+1) is the> positive square root of 2. It's clear that for
all n there exists a j> such that c(j)_i = b_i for i in 1...n,
but b is not any of the c(j).> So if one believes that proving
this theorem about some countable list> C of numbers and some
number b suggests that b is in the list C, one> has to think
that the square root of two is rational (indeed one has> to
think that all real numbers are rational, since the abilty to
have> a (not necessarily terminating) expansion in any base
from 2 upwards> is a property of all real numbers.> M.I
outlined this problem with Pi and the rationals with the
DiagNotEqual function.It appears when :b = IRRATIONALSc =
RATIONALSI'm forced to take a Godelian stance here. Where b =
DIAG, b is clearlycomputable in that we can calculate its
digits ad infinitum, yet b does notappear on the list of
computables, (for infinite digits).If b was formulated by an
algorithm it would appear in the preliminary gridUTM(x,y), but
when x = y it would recurse and fail to terminate, and
neverqualify as effectively computable.Or another function at
c(j) could be used that outputs 0. b_1 ... b_j-1 0 0 0 0So
just like the Godel statement is True but doesn't belong to
the list ofproven statements, DIAG is a computable number but
doesn't belong to the listof computable numbers.A direct
mathematical analogy might correspond between the
diagonalisationtechniques used in both.Proving computable
numbers are not countable is the last thing I wanted to do
:-)Herc===
===
Subject:
: Re: Cantor's Diagonal ArgumentIn sci.logic,
|-|erc<contactvia@wwwadamskingdom.com><4054fa24$0$8357$afc38c87
@news.optusnet.com.au>:> Ouch!! Doesn't this mean b = c(j)?
and is it correctly derived?> for all n, exists j, b_1 =
c(j)_1, b_2 = c(j)_2, ... b_n = c(j)_n> This is equivalent
to : for all n, b_n = c(j)_n> You have to prove that there
is such a j.> note the exists j> for all n, exists j, b_1 =
c(j)_1, b_2 = c(j)_2, ... b_n = c(j)_n> admitedly this is
different to :> exists j, for all n, b_n = c(j)_n> I showed
that no matter how many digits you examine, the diag>> up to
that many digits is covered on the list.> Herc>>You have to
show thatfor all n, there exists a j, such that for all i,>>
b_n_i = c(j)_i> b_n is a particular b? I'll write b(n) and
keep _i notation for digit indexing.> I have :> for all n, for
all i, exists j> b(n)_i = c(j)_i> You want :> for all n, exists
j, for all i> b(n)_i = c(j)_i> where> b = DIAG> c =
COMPUTABLESThat's right, and they're different.>> Yes, that's
what he would need to show.>> It might be helpful though to
express what he has shown in terms of>> what sort of nunmbers
are involved, because it's often easier to see>> things
expressed in simple categories instead of in something like>>
predicate calculus. And to show how the same thing can be
proved>> about simpler lists and rather more obvious
non-members.>> So what does for all n, exists j, b_i = c)j)_i
for i in 1....n mean.>> It means If x is the number
represented by b_1.....b_n, then some>> number whose
expression has those n digits as the first n is on the>> list.
If the list includes all numbers whose expansion terminates>>
after a finite number of terms this is trivially true, since x
itself>> is on the list. Now a list of all rational numbers
certainly includes>> all numbers with finite expansions, so
it's clear that if the list is>> the list of all rational
numbers and b (represented as b_1....) is any>> real number
then for any n we can find the number b_1....b_n on the>>
list.>> Think about the well known fact, established by the
ancient Greeks,>> that there is no rational number whose
square is 2. Mow let C(j) be>> the complete set of rationals
between 0 and 1 indexed by natural>> numbers j, and let b be
the real number such that (b+1) is the>> positive square root
of 2. It's clear that for all n there exists a j>> such that
c(j)_i = b_i for i in 1...n, but b is not any of the c(j).>>
So if one believes that proving this theorem about some
countable list>> C of numbers and some number b suggests that
b is in the list C, one>> has to think that the square root of
two is rational (indeed one has>> to think that all real
numbers are rational, since the abilty to have>> a (not
necessarily terminating) expansion in any base from 2
upwards>> is a property of all real numbers.>> M.> I outlined
this problem with Pi and the rationals with the> DiagNotEqual
function.> It appears when :> b = IRRATIONALS> c = RATIONALS>
I'm forced to take a Godelian stance here. Where b = DIAG, b
is clearly> computable in that we can calculate its digits ad
infinitum, yet b does not> appear on the list of computables,
(for infinite digits).> If b was formulated by an algorithm it
would appear in the preliminary grid> UTM(x,y), but when x = y
it would recurse and fail to terminate, and never> qualify as
effectively computable.> Or another function at c(j) could be
used that outputs 0. b_1 ... b_j-1 0 0 0 0> So just like the
Godel statement is True but doesn't belong to the list of>
proven statements, DIAG is a computable number but doesn't
belong to the list> of computable numbers.> A direct
mathematical analogy might correspond between the
diagonalisation> techniques used in both.> Proving computable
numbers are not countable is the last thing I> wanted to do
:-)The computable numbers are countable. The real numbers are
not.> Herc-- #191, ewill3@earthlink.netIt's still legal to go
.sigless.===
===
Subject:
: Re: Cantor's Diagonal Argument> I have :
for all n, for all i, exists j> b(n)_i = c(j)_i& b(n)_1 =
c(j)_1& b(n)_2 = c(j)_2up to _i You want : for all n, exists
j, for all i> b(n)_i = c(j)_i where> b = DIAG> c =
COMPUTABLES> That's right, and they're different.agreed, but
is it significant?they both mean the number is on the list to
any degree required.the 1st digit is on the list,and the 2nd
digit is on the list too,and the 3rd digit is on the list
too....all digits are on the listtherefore the number is on
the list.> The computable numbers are countable. The real
numbers are not.fantacy, non computable number is a
misnomer.Herc===
===
Subject:
: Re: Cantor's Diagonal Argument> I
have :> for all n, for all i, exists j> b(n)_i = c(j)_i> &
b(n)_1 = c(j)_1> & b(n)_2 = c(j)_2> up to _i> You want :>
for all n, exists j, for all i> b(n)_i = c(j)_i> where> b =
DIAG> c = COMPUTABLES That's right, and they're different.
agreed, but is it significant?> they both mean the number is
on the list to any degree required.> the 1st digit is on the
list,> and the 2nd digit is on the list too,> and the 3rd
digit is on the list too> ....> all digits are on the list>
therefore the number is on the list.I'll introduce a
functionEquals(DE1, DE2, n) { if (n = 0) { Return TRUE }
elseif (DE1_n = DE2_n) { Equals(DE1, DE2, n-1) } else { Return
FALSE }}Forall n, Equals(DE1, DE2, n) -> DE1 = DE2If extended
to handle recurring 9s,does this satisfy the accepted
definition of = ?Herc===
===
Subject:
: Find the third point of a
triangleI have a Traingle ABC. I have the position of two of
the points say A(x1,y1) and B (x2,y2). I also have the length
of line segments AB andBC.Q) How do I find the position of the
third point on a 2D plane, inthis case C ?Tejas===
===
Subject:
: Re:
Find the third point of a triangle> I have a Traingle ABC. I
have the position of two of the points say A> (x1,y1) and B
(x2,y2). I also have the length of line segments AB and> BC.>
Q) How do I find the position of the third point on a 2D
plane, in> this case C ?How many answers would you
like?===
===
Subject:
: Re: Find the third point of a triangle*
tejasm@hotmail.comI have a Traingle ABC. I have the position
of two of the points say A> (x1,y1) and B (x2,y2). I also have
the length of line segments AB and> BC.Q) How do I find the
position of the third point on a 2D plane, in> this case C ?Is
this homework? If so, please tell us so. If not, I apoligize,
butstill, please tell us that too.Anyway, consider an easier
case, where (x1,y1)=(0,0) and(x2,y2)=(1,0), and AB and BC is u
and v respectively. I am sure youcan, by using Pythagoras, come
up with two equations involving thethird point (x3,y3) and the
lengths u and v.Two equations with two unknowns are
solvable.After that, generalize.-- Jon Haugsand Dept. of
Informatics, Univ. of Oslo, Norway, mailto:jonhaug@ifi.uio.no
http://www.ifi.uio.no/~jonhaug/, Phone: +47 22 85 24
92===
===
Subject:
: re:Cantor Paradox :-)> The problem is evidently
on your end, it seems.Unfortunately I don't have any idea what
to do about it, esp.considering the fact that I'm not even
aware of the problem exceptfrom your complaints.Unless I state
explicitly otherwise (by quoting a paragraph), myreplies
address to the latest post present at the moment I hit
thereply button. ===
===
Subject:
: Re: Cantor Paradox :-)> What do
you mean by provable that E!yPy and Px hold? Does the>
provable extend to Px, so that Px is provable? Or do you mean>
that P is true of x? If the former, what formula are you
talking> about?Of course I mean that Px is provable and also
that it's provable thatx is real. ===
===
Subject:
: Re: Cantor
Paradox :-)Theorist says...>> What do you mean by provable
that E!yPy and Px hold? Does the>> provable extend to Px, so
that Px is provable? Or do you mean>> that P is true of x? If
the former, what formula are you talking>> about?>Of course I
mean that Px is provable and also that it's provable that>x is
real.So you mean something like this: { <P,x> | ZFC proves
(P(x) & E!y (P(y))) }But that doesn't work. Note that 'x' is a
*free* variable in thesentence (P(x) & E!y (P(y))). It is a
theorem of most formulationsof first order-logic that if you
can prove a statement with a freevariable, then you can prove
the universally quantified formula. Butyou certainly don't
mean to say { <P,x> | ZFC proves forall x, (P(x) & E!y (P(y)))
}The problem is that the x appearing in <P,x> is unconnectedto
the x appearing in 'ZFC proves forall x, (P(x) & E!y
(P(y)))'.What you really want to do is to have not a single
formula(P(x) & E!y (P(y))) but a bunch of *different*
formulas, onefor each real number r. That would be possible if
there werea constant symbol for each real number r. Then you
could talkabout the formula that results from replacing 'x'
in(P(x) & E!y (P(y))) by the constant symbol for r. But ZFC
doesn'thave constant symbols for each r.You could certainly
imagine extending ZFC by *adding* a constantsymbol for each r.
But that requires an uncountable number of
constantsymbols.--Daryl McCulloughIthaca, NY ===
===
Subject:
: Re:
Cantor Paradox :-)> As far as that goes, I am naive enough to
believe that when you> develop intuitions about these matters,
the distinctions become> clearer not cloudier. Heck, I'm even
naive enough to believe thatI> have some intuition about those
stuff.> At some point you stop paying attention to those
details. As long asyour main ideas are correct (in this case
my main idea was that thetruth of a statement has a very mild
relation to its form and thusit's useless for defining a list
of all definable reals) and you arecapable to fill in the
details accurately if necessary it never leadsto disaster.> By
the way, since your posting software is borked and won't put
yourposts in the proper place in the thread (because it
dropsreferences),> it would be especially helpful if you
included some context whenyou> follow up. Otherwise, we have
to guess who you're responding to.Actually I've never seen
those problems you're talking about on any ofthe computers I
viewed this forum from. So it might be your softwarethat's
incompatible. But for the convenience of you and other
whomight experience problems with my posts I'm gonna put
context fromnow on. ===
===
Subject:
: Re: Cantor Paradox :-)
Discussion, linux)>> As far as that goes, I am naive enough to
believe that when you>> develop intuitions about these matters,
the distinctions become>> clearer not cloudier. Heck, I'm even
naive enough to believe that> I>> have some intuition about
those stuff.>At some point you stop paying attention to those
details. As long as> your main ideas are correct (in this case
my main idea was that the> truth of a statement has a very mild
relation to its form and thus> it's useless for defining a list
of all definable reals) and you are> capable to fill in the
details accurately if necessary it never leads> to
disaster.Not disaster, but it does lead to long, dull
exchanges that could beavoided if you kept the concepts clear
and distinct and expressed theissues properly. In my ever so
humble opinion.>> By the way, since your posting software is
borked and won't put your> posts in the proper place in the
thread (because it drops> references),>> it would be
especially helpful if you included some context when> you>>
follow up. Otherwise, we have to guess who you're responding
to.> Actually I've never seen those problems you're talking
about on any of> the computers I viewed this forum from. So it
might be your software> that's incompatible. But for the
convenience of you and other who> might experience problems
with my posts I'm gonna put context from> now on.I appreciate
the concession, but it's not my software. Here's
yourreferences header.That's the full header. One reference,
to the top of the thread.Now, David Ullrich's recent
experiences lead one to wonder: could itbe a problem with my
newsserver or even some upstream server (doingsomething
dastardly to the References line)? The answer is: evidentlynot
-- or the culprit is upstream of Google Groups, too.
GoogleGroups clearly shows every single one of your posts in
this threaddirectly under the root of the thread. Each of them
has a referencesheader with just one entry, and that entry
points to the root of thethread, no matter who you're
following up.The problem is evidently on your end, it seems.--
And hey, if you're moping and miserable because mathematics
tests you,then maybe, if you think you're a mathematician, you
might want to trya different field. -- Another James S. Harris
self-diagnosis.===
===
Subject:
: Re: Cantor Paradox :-)Your argument
is obscure. What does it look like, stated in full?I take the
set of all pairs (P, x) where P is a first order formulawith
one free variable and x real so that it's provable that
E!yPyand Px hold.Suppose that the consistency of the system
we're considering (whichcontains ZF) is provable in that
system.Now I prove that {(P, x)} is a function. If it's not,
there are x1, x2and P so that Px1, Px2 and E!yPy are provable,
which contradictsconsistency.Therefore it is a function. Now I
take some effective bijectionbetween {P| E!yPy is provable}
and the natural numbers (it's easy tosee that it exists) and
use the function {(P, x)} to effectivelyenlist all the reals
definable in this sense.Now I take the (anti)diagonal and see
that it's definable as well butnot on the list.This is a
contradiction. ===
===
Subject:
: Re: Cantor Paradox :-)> I take the
set of all pairs (P, x) where P is a first order formula> with
one free variable and x real so that it's provable that E!yPy>
and Px hold. What do you mean by provable that E!yPy and Px
hold? Does theprovable extend to Px, so that Px is provable?
Or do you meanthat P is true of x? If the former, what formula
are you talkingabout? ===
===
Subject:
: re:Cantor Paradox :-)That
depends on where you're coming from. For some
peoplecomputability is the first thing that comes to mind when
they hearabout defining a list.Especially considering the fact
that the most important property ofrecursive functions is the
lack of an effectively computable list ofthem.It's not
cruicial whether the truth value discussed above
isrepresentable or computable or whatever, because it's
equallyunlikely that it is.I apologize for using the wrong
word, but if you develop someintuition about those stuff,
you'll see there isn't much of adifference. ===
===
Subject:
: Re:
Cantor Paradox :-) Discussion, linux)> That depends on where
you're coming from. For some people> computability is the
first thing that comes to mind when they hear> about defining
a list.But that's a limitation of some people, I guess, and
not of Nathan'sargument[1]. The context was nominally Cantor's
setting (althoughNathan's argument has no evident implication
for Cantor's theorem), sothere's no reason to think that it's
a computability issue.> It's not cruicial whether the truth
value discussed above is> representable or computable or
whatever, because it's equally> unlikely that it is.> I
apologize for using the wrong word, but if you develop some>
intuition about those stuff, you'll see there isn't much of a>
difference.There is a difference. The predicate machine n halts
on input n isrepresentable but not recursive. As far as that
goes, I am naive enough to believe that when youdevelop
intuitions about these matters, the distinctions becomeclearer
not cloudier. Heck, I'm even naive enough to believe that Ihave
some intuition about those stuff.By the way, since your posting
software is borked and won't put yourposts in the proper place
in the thread (because it drops references),it would be
especially helpful if you included some context when youfollow
up. Otherwise, we have to guess who you're responding
to.Footnotes: [1] Really, we left Nathan's argument when we
moved to formallanguages, but as has been pointed out,
Nathan's argument wasn'tNathan's anyway.-- And besides, you
lied, it's by *convention* that [X], not bydefinition. --
James Harris, on the subtleties of conventions.===
===
Subject:
: Re:
Cantor Paradox :-) You have to a be mathematician. You circular
> logic is actually 2 stupid for a logician.Bzzt. Not this
week. This week, I am a osopher. That's physically impossible.
The difference between a osopher and a osopher-of-mathematics,
is that osophers-of-mathematics have computers that go Bzzt,
and osophers have computers that go clunk.===
===
Subject:
: Re:
Cantor Paradox :-) <87ishcgc0s.fsf@phiwumbda.org>
<e4a0829b.0403101303.38a02b85@posting.google.com>
<87ptbkcrrq.fsf@phiwumbda.org>
<e4a0829b.0403102111.34217e33@posting.google.com>
<87brn3dhzt.fsf@phiwumbda.org>
<e4a0829b.0403110415.1e89770@posting.google.com>
<87smgfbcn0.fsf@phiwumbda.org>
<e4a0829b.0403150828.19f2ec00@posting.google.com> Discussion,
linux)>>> You have to a be mathematician. You circular >>
logic is actually 2 stupid for a logician.>>> Bzzt. Not this
week. This week, I am a osopher.> That's physically impossible.
The difference> between a osopher and a
osopher-of-mathematics,> is that osophers-of-mathematics have
computers> that go Bzzt, and osophers have computers> that go
clunk.Sorry, osopher-of-mathematics isn't in my current
jobdescription. Hence, my computer goes clunk, not Bzzt.On the
other hand, *I* sometimes go Bzzt. That must be what
caused(part of) your confusion.-- What I've learned is that
[mathematicians are] the gatekeepers, andseem to have almost
absolute power when it comes to mathematics. -- James Harris,
on All I Really Ever Needed to Know I Learned in
/Ghostbusters/.===
===
Subject:
: Re: Cantor Paradox :-)>>> You
have to a be mathematician. You circular >> logic is actually
2 stupid for a logician.>>> Bzzt. Not this week. This week,
I am a osopher. That's physically impossible. The difference>
between a osopher and a osopher-of-mathematics,> is that
osophers-of-mathematics have computers> that go Bzzt, and
osophers have computers> that go clunk.Sorry,
osopher-of-mathematics isn't in my current job> description.
Hence, my computer goes clunk, not Bzzt. It's doesn't matter
what your job description is, since mathematicians invented
evolution, they have auto-clunker computers. On the other
hand, *I* sometimes go Bzzt. That must be what caused> (part
of) your confusion. The whole part of your confusion is that
you think other people actually read Kant .===
===
Subject:
:
re:Cantor Paradox :-)Representability and computability are
very close notions. Since theOP didn't specify what he assumed
to produce his list saying that hecan't assume computability is
a reasonable objection to hisargument.Nevertheless
representability would be a more appropriate notion touse.My
argument still works though and establishes three facts:1) The
truth of a formula is irrepresentable (Tarski's theorem
forZF).2) Consistensy of ZF or any stronger (but countable)
system isundecidable.3) The OP's apparent attempt to deduce a
contradiction from ZF doesn'twork. ===
===
Subject:
: Re: Cantor
Paradox :-) Discussion, linux)> Representability and
computability are very close notions. Since the> OP didn't
specify what he assumed to produce his list saying that he>
can't assume computability is a reasonable objection to his>
argument.Sorry, but that's just nonsense.He didn't say whether
he assumed that the list was in ascending ordereither, but
saying that he can't assume it is in ascending order isn'ta
reasonable objection.He never used the assumption (not even
implicitly) that the list wascomputable. He never even
mentioned computability. To say thattherefore, it's relevant
and reasonable to object that it isn'tcomputable is, well,
odd.-- Jesse HughesShe moaned, in pain and pleasure, as, in a
confused whirlwind, sheglimpsed an image of Saint Sebastian
riddled with arrows, crucifiedand impaled. --Mario Vargas
Llosa on category theory===
===
Subject:
: Re: Cantor Paradox :-)My
argument still works though and establishes three facts:> 1)
The truth of a formula is irrepresentable (Tarski's theorem
for> ZF).2) Consistensy of ZF or any stronger (but countable)
system is> undecidable. What argument for 2) do you have in
mind?===
===
Subject:
: Re: Cantor Paradox :-)> This can be used to
show Godel's second incompleteness theorem for ZF Your
argument is obscure. What does it look like, stated in
full?===
===
Subject:
: Re: Cantor Paradox :-)It's also a set if E!x
P(x) is not recursive. Honest. Check the> axioms of ZFC. See
whether or not the separation axiom mentionsthat> the formula
must be recursive.> Forget about recursivity for a moment.A
formula P is just a string of characters. There's no way to
expressits truth value from its form by a first order formula.
That truthvalue is not even computable in the weak sense
required by theseperation axiom.Not only it's incomputable,
for some P it depends on the model of ZFwe're
considering.Whether you see computability as relevent or not
is irrelevant.Fact is that the diagonal can't be constructed
either approach youuse. ===
===
Subject:
: Re: Cantor Paradox :-)
Discussion, linux)>>> It's also a set if E!x P(x) is not
recursive. Honest. Check the>> axioms of ZFC. See whether or
not the separation axiom mentions> that>> the formula must be
recursive.>Forget about recursivity for a moment.> A formula P
is just a string of characters. There's no way to express> its
truth value from its form by a first order formula. That
truth> value is not even computable in the weak sense required
by the> seperation axiom.If you'll drop the word computable in
favor of representable, thenI'll agree. I thought about things
this morning and realized that a million or sotimes, I had said
that the set in question is well-defined. It'swell-defined only
if the predicate n encodes P and |= E!x P(x) isrepresentable.
But that predicate isn't representable. I was wrongabout
that.Nonetheless, I think your use of the word computable even
in thispost is distracting and unfortunate. Representable is
more to thepoint. I wish I remembered the details of
representability betterthan I do, but I don't, so the less I
say on the matter the better.> Not only it's incomputable, for
some P it depends on the model of ZF> we're considering.>
Whether you see computability as relevent or not is
irrelevant.> Fact is that the diagonal can't be constructed
either approach you> use.Well, on this point we've never
disagreed and I've never assertedotherwise.-- These
mathematicians are worse than communists, as how do you
explaintheir behavior? I *am* the American Dream, fighting for
what should bemine, having to get past weak-minded academics
who are fighting toblock my success. But I shall prevail!!! --
James S. Harris===
===
Subject:
: re:puzzle: GCDs of Infinite Set of
Integer PairsIf you assume the constructibility axiom (V=L)
then the entire classof all sets can be constructively well
ordered. So I wonder what thatresult could be. ===
===
Subject:
: Re:
How to solve a polynomial of degree 6 over a ring modulo a
composite by support1.mathforum.org (8.11.6/8.11.6/The Math
Forum, $Revision: 1.9 primary) id i2FMNu117529;>Mukesh Kumar
Singh schrieb:>>> Dear Research Community,>> I have a
generic polynomial equation..>>> f(t)= t^6 + a. t^5 + b. t^4
+ c. t^3 + d. t^2 + e. t + g = 0>>> I need to solve the above
polynomial of degree 6 over a ring modulo a>> composite n (n is
large of the order of 1024 bbits). Where the factors>> of n is
unknown and it is hard to find too. I want a solution that>>
does not amount to factoring n.>> I have a finite but large
enough set of {a,b,c,d,e,g} for which I need>> to know at
least one solution.>I would try (to modify) the algorithm for
roots mod prime (e.g. Henry Cohen,>Springer GTM 138, Algo
1.6.1, p.37 or Knuth TAOCP vol. 2). You must check>whether a
division by zero occurs (i.e. you found a factor of n). (sorry
if this>is too naive).>Contrary to the reply in sci.math>it
seems that the solution of a polynomial mod n does not
necessarily imply a>factorization of n.>Trivial example>Factor
f(x)=x^2-4 mod n =(x-2)(x+2)>A road to less trivial examples:
Fix n s.t. 2^1024< n <2^1025 and choose, e.g. q>ca.
2^513.>f(x)=(x-q)^2-4>can still be easily factored (see
Cohen), but it is quite unlikely to find a>factor of n in
doing so.>hth>Klaus>> Can anybody please help?>> thanks in
advance..>> regards,>> MukeshNonetheless, the ability to
factor a generic degree-6 polynomial mod n implies the ability
to factor n. The usual way to prove this:Pick a1,a2,...,a6 in
Z/n at random.Form the polynomial (x-a1)(x-a2)...(x-a6) over
Z/n.Give it to your polynomial-factoring oracle.If he extracts
a linear factor (x-b1), take gcd(b1-ai,n), i=1,2,..,6.(If p and
q are two of the factors of n, we will have b1=ai(mod p)and
b1=aj(mod q), but there is no reason for i=j.)If he extracts a
quadratic factor (x^2+k*x+m), take the pairs (ai,aj)of original
roots, form quadratics (x-ai)(x-aj), and compare
thecoefficients (-ai-aj, ai*aj) to the coefficients (k,m) via
gcd.Again, usually p will fall out as
gcd(-ai-aj-k,n).===
===
Subject:
: Re: gamma function> Since there's
not good way to put this in terms of ASCII, I'll use> page
references. How would one go from equation #2 to equation #3
on> http://mathworld.wolfram.com/GammaFunction.html ? I keep
thinking> that it's either substitution or integration by
parts, but I can't> seem to find anything that works.Should be
easy : substitute T=t^2 .Then dT=2t dt. What is giving
youtrouble?> Thank you,> Colvin===
===
Subject:
: Re: Sum of Product
= Product of Sum (number-theoretical)> I will post here a
generalization of the result in Sum Of Powers Of>
Divisor-Function.>
(http://groups.google.com/groups?dq=&start=50&hl=en&lr=&ie=UTF
-8&group=sci.math&safe=off&selm=b4be2fdf.0403131607.46772492%
40posting.google.com)Let a(p,k) be a nonnegative integer such
that> p^a(p,k) is the highest power of the prime p which
divides k.> Let {b(k)} be any sequence defined for all
positive integers k.> Then, for every positive integer m:---
---> | |> / ( | | b(a(p,k)) )> --- p|k> k|m => a(p,m)> ----
--> ! | > ! | (1 + / b(k) )> p|m --> k=1> (All p-products in
this post are over primes p which divide k or m,>
depending.)In linear-mode:sum{k|m} (product{p|k} b(a(p,k)) )
=product{p|m} (1 +sum{k=1 to a(p,m)} b(k) ).And from this we
can get, for any positive integer n:--- --- --- ---> | |> / /
... / ( | | b(a(p,k_n)) )> --- --- --- p|k_n> k_1|m k_2|k_1
k_n|k_{n-1}=> a(p,m) k_1 k_{n-1}> ---- -- -- --> ! |
/a(p,m)+n-1 > ! | (| | + / / ... / b(k_n) )> p|m n - 1 / -- --
--> k_1=1 k_2=1 k_n=1> In
linear-mode:sum{k_1|m}sum{k_2|k_1}...> sum{k_n|k_{n-1}}
(product{p|k_n} b(a(p,k_n)) ) =product{p|m}
(binomial(a(p,m)+n-1,n-1) +> +sum{k_1=1 to a(p,m)}sum{k_2=1 to
k_1}...> sum{k_n=1 to k_{n-1}} b(k_n) )> If we define b(0) = 1,
then we can have the simpler:! ! --- --- --- ---! | |! / / ...
/ ( | | b(a(p,k_n)) )! --- --- --- p|k_n! k_1|m k_2|k_1
k_n|k_{n-1}! ! =! a(p,m) k_1 k_{n-1}! ---- -- -- --! ! | ! ! |
(/ / ... / b(k_n) )! p|m -- -- --! k_1=0 k_2=0 k_n=0! In
linear-mode: sum{k_1|m}sum{k_2|k_1}... sum{k_n|k_{n-1}}
(product{p|k_n} b(a(p,k_n)) ) = product{p|m} (sum{k_1=0 to
a(p,m)}sum{k_2=0 to k_1}... sum{k_n=0 to k_{n-1}} b(k_n) )
Leroy > For n = 2 we also have simply> --- ---> | |> / d(m/k)
( | | b(a(p,k)) )> --- p|k> k|m => a(p,m) k_1> ---- -- -- > !
| > ! | (a(p,m) + 1 + / / b(k_2) )> p|m -- --> k_1=1 k_2=1>
d() is the number-of-divisors function.> In
linear-mode:sum{k|m} d(m/k) (product{p|k} b(a(p,k)) )
=product{p|m} (a(p,m) +1 +> +sum{k_1=1 to a(p,m)}sum{k_2=1 to
k_1} b(k_2) ) > So, an interesting specific case of the
above:sum{k|m} d(m/k) /d(k) => product{p|m} ((a(p,m) +
2)(H(a(p,m)+2) -1)),> where H(j) = sum{k=1 to j} 1/k,> the
j_th harmonic number.> A specific example of first identity:>
sum{k|m} product{p|k} F(a(p,k))=product{p|m} F(a(p,m)+2),where
F(j) is the j_th Fibonacci number.> Leroy Quet===
===
Subject:
: Re:
Sum Of Powers Of Divisor-Function> First, some specific
identities, then the general result, then some limits.Each of
these is probably very well known, or at least easily found.>
But they seem very deserving of being even more well known, so
I will post these.Let d(n) be the number of positive divisors
of n.Then, for m = any positive integer:> sum{k|m^2} d(k) =
d(m) d(m^2).> sum{k|m^6} d(k)^2 = d(m^3) d(m^4) d(m^6).>
sum{k|m^2} d(k)^3 = d(m)^2 d(m^2)^2.> I wonder what is known
about such identities as the above.Are there any other
interesting sum-identities like the above fordifferent
exponents?Leroy Quet -Generally,sum{k|m} d(k)^r
=product{p=primes,p|m} (sum{k=1 to a(p,k)+1} k^r),where
p^a(p,k) = highest power of prime p which divides k.-2
limits:For r > 1,limit{n-> oo} sum{k|m^n} 1/d(k)^r
=zeta(r)^b(m),where b(m) is the number of distinct primes
dividing m.For r >= 0,limit{n-> oo}(sum{k|m^n} d(k)^r
)/(n^{(r+1)b(m)}) = (product{p|m} a(p,m)^(r+1))/(r+1)^b(m).> I
may have made an error, but the above results are easily found
(if right).> Leroy Quet===
===
Subject:
: Help: Numerically filtering
wave forms from SignalsHi All,I'm interested in how the best
way to numerically filter a waveform ofa specific period from
a signal. For example(1) Consider we were looking at the
temperature in a London as afunction of time. This may be
thought of as a superpositioning ofwave forms of various
frequencies (I think this was part of Fourier'sTheorem).
Clearly there is a substantial contribution from a wave
ofperiod one day. If T(t) is our temperature in London at time
t, howdo we numerically find the wave of 24 hour period W24(t)
and theremainder function T(t) such that T(t) = W24(t) +
R(t)(1) Say we had a stock price as a function of time. Say
daily closeprice over 10 years P(t). If we believed there was
a yearly cycle init, how would we find this wave form and a
remainder function similarto above.Hope someone can shed some
light on it. Best I have been able to comeup with is that a
simple moving average of length T actually filtersout (very
roughly) a wave of period T and its harmonics. Hence thesignal
less SMA(T) actually gives a very rough approximation of a
waveform of period T. Even better is SMA(T) - SMA(2T) to
roughly loosethe harmonics.Any takers?Stewart===
===
Subject:
: Re:
Integer-Alterations On a Grid (fun)One final reply to this
thread: > It seems that this game's strategy would rely
heavily upon the actual> integer-alteration rules being used
for any particular round.> Knowing what you know about the
general game, what would be some of> YOUR suggestions for
integer-alteration rules which would make for> interesting
play and strategy?> ;)> Included among the rules can be those
which give an output as a result> of an> if/then question,
such as: m = m+3 if m is a prime; m = m^2 otherwise. >
Possibly you could have rules which vary m based on the other
player'srules in the columns, if your rules are in the rows,
and vice versa.Such as, for row-rule:m = value attained by
square's column-rule + 1(or something more creative)You CAN
have rules which give you a range of possible values, andwhich
value is used is determined by the player.But, remember, both
players are subject to the same rules. So anadvantage to you
is also an advantage for your opponent.(I personally suggest
keeping the rules relatively simple. But havinga given game
where the rules are as complicated as possible could befun
too.)> But I would discourage such rules which pretty much
give the same> output no matter what is inputted, such as 'm =
4' or anything less> obviously of one-output. (Even such rules
as m = floor(|m|^(1/|m|)),> which only outputs 1 unless m is
0, should probably be discouraged im> most cases.)As far as m
= floor(|m|^(1/|m|)), since no player can land on asquare with
an m of 0, this rule is equivalent to m = 1.Remember:> Goal of
game = FUN!Leroy QuetHas anyone played the game?Leroy Quet>
This is a game I made up today.> It is unoriginal, in that,
not only do I steal ideas from my own games I > have posted
here in the past, but from other people's games also.> But,
anyway, this game seems as if it would be SO fun to play that
I HAD > to post it!> 2 players.> Played on n-by-n grid (as
many of my games are).> One player makes up n
integer-alteration rules (see below for examples) > for the n
rows, one rule per row.> And the other player makes up n rules
for the n columns, one rule per > column.> Rule: a
(meta)mathematical rule which changes any input integer into
an > output integer.> At the game's beginning, after the rules
are created, each player gets 3 > random integers, from 1 to
n.> A player's first 2 random integer give the coordinate of
the square the > player starts in, the last integer gives the
player's starting integer.> And what player moves first is
then chosen randomly.> (But no more randomnes is in the game
once play begins.)> At each turn, a player can choose to apply
the row-rule OR the > column-rule to his/her integer. (Each
player has their own integer which > is altered.)> the square
he/she is currently at.> The player can then choose to move
either up, down, left, or right, > without regard to what
integers (s)he must jump over, that number (m) > of
grid-squares.> And the grid is considered topologically a
torus, ie. going off the right > edge, for example, brings you
back to the left side> (and the same (mod n)-topology for
right-to-left or vertical moves).> (So, for example, a move
from column 3 to the right 7 squares, on an n = > 6 grid, gets
the player to column 2, which is one square to the *left* of >
where (s)he started.)> But a player must only move to a square
*without* an integer (written > there by either player) already
in it.> And the game's winner is the last player able to move.>
Notes:> *Be Creative in coming up with rules!> *The output
integers' signs are, in practice, insignificant, since you >
can move right OR left, up OR down. But an output of 0 is to
be avoided!> *Start out with easy rules as you first learn how
to play this game, or > if you simply are poor at math. It
would be less fun to play this game if > you must spend an
hour calculating each of your integers!> Example of rules:>
(For n=6)> Rows:> A) m = m+1;> B) m = 2*m;> C) m = m + sum of
distinct primes dividing all already-written > neighboring
integers;> D) m = d(|m|)*phi(|m|) -m; (d(m) is
number-of-divisors function, phi(m) > is Euler-phi function.)>
E) m = (numerator)-(denominator) of reduced rational,> sum{k=1
to |m|} 1/k;> F) m = m + number of already-written integers in
row F.> Columns:> 1) m = ceiling(sqrt(|m|))> 2) m =
ceiling(secant(m)), m in radians;> 3) m = number of primes <=
|m|;> 4) m = 2^|m| - m;> 5) m = number of divisors in product
of all aready-written integers in > column 1;> 6) m =
ceiling((|m|th Fibonacci number)/(|m|_th prime)).> (Yes,
computers might be needed for some kinds of rules, and are
allowed.)> And a grid to aid visualization:>
------------------> A! ! ! ! ! ! !> ------------------> B! ! !
! ! ! !> ------------------> C! ! ! ! ! ! !>
------------------> D! ! ! ! ! ! !> ------------------> E! ! !
! ! ! !> ------------------> F! ! ! ! ! ! !>
------------------> 1 2 3 4 5 6> Leroy Quet===
===
Subject:
: Re:
Copyrights, fair use, and Internet realities... by doing a
Google search on *my* name and math.> Hey, look what shows up
when I do a search on *your* name and
crank:http://www.crank.net/harris.html===
===
Subject:
: Re:
Copyrights, fair use, and Internet realities... > A person
like Dik Winter needs to copy my work because he can't put up
> something of his own to draw attention, or even if he can,
he STILL > feels a need to try and draw attention to himself
using me.Interesting reasoning. More so because before you
drew attention tothat page the page was not even in the top 25
of my pages. Yes, itnow has plummeted to 15th place. I will
look what it will do in thecourse of this month.-- dik t.
winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland,
+31205924131home: bovenover 215, 1025 jn amsterdam, nederland;
http://www.cwi.nl/~dik/===
===
Subject:
: Re: Copyrights, fair use,
and Internet realities>>I noticed Winter's pages by doing a
Google search on *my* name and math.Searching google for
yourself? That's slightly pathetic.Searching for oneself on
Google isn't pathetic for people that have websites or have
participated on usenet or on other online forums using
theirreal name.It's useful to know what kind of stuff,
especially personal information,about you is easily findable.
A lot of people are surprised when they findout, for example,
that by typing their name and their city into Google, itcomes
up with their address, phone number, and a link to a map of
thataddress.-- --Tim Smith===
===
Subject:
: Re: Copyrights, fair
use, and Internet realities> Some of you may think it's ok for
Dik Winter to copy from a post I> made to Usenet and put that
on a webpage without my permission and> keep it up despite my
protests.>[snip misguided whining]You posted something to
USENET. People have already explained how it works to you, so
I won't repeat them.He then took what you posted, and created
a webpage detailing what is wrong with your claim. He did this
because your reasoning was incorrect, not to make himself
money. If you really cared about the math you're trying to
peddle, you would be glad to have someone correct any mistake
you may have made. If you aren't, then I suppose you are more
interested in puffing up your ego than getting the math
right.-Agent1===
===
Subject:
: Re: Copyrights, fair use, and
Internet realities> One of my pages:
<http://csf.colorado.edu/pkt/pktauthors/Vienneau.Robert/
LTV-FAQ.html>A copy that was made without informing me:
<http://www.simons-rock.edu/~eatonak/LTV-FAQ.html>I have no
objection.> That is not, however, copying a usenet post that
you made. That's copying a website that you made. There is a
bit of a difference.-- email: wtwentyman at copper dot
net===
===
Subject:
: Double-Sum Over Coprime Indexes (puzzle
problem)I bet this is one of the easiest math puzzle, but it
seems worth posting anyway.Evaluate: oo oo--- --- 1
-----------/ / k j (k+j)--- ---k=1 j=1GCD(k,j)=1In linear
mode:sum{k=1 to oo} sum {j=1 to oo, GCD(k,j)=1}
1/(kj(k+j)).Leroy Quet===
===
Subject:
: locally compact space with
an intrinsic metricAn intrinsic metric is defined to be one
such that for any two pointsx and y, there is a third point z
such that d(x,z)=d(y,z)=1/2 d(x,y).In the book I'm reading it
says that in a locally compact space withan intrinsic metric,
every bounded and closed set is compact. Doesanyone know how
to prove that?===
===
Subject:
: Re: locally compact space with an
intrinsic metric> An intrinsic metric is defined to be one
such that for any two points> x and y, there is a third point
z such that d(x,z)=d(y,z)=1/2 d(x,y).In the book I'm reading
it says that in a locally compact space with> an intrinsic
metric, every bounded and closed set is compact. Does> anyone
know how to prove that?(0,1) with the usual metric would
appear to be a counterexample.===
===
Subject:
: Re: Copyright Issue:
Dik Winter publishing my work> What can I do...?> Well, any
normal usenet kook would assume multiple identities andbombard
the newsgroup with outrageous claims. Then after a few
weeks,you should suddenly disappear. That is what I expect you
should do.===
===
Subject:
: Re: Copyright Issue: Dik Winter
publishing my workIn sci.math.num-analysis, James
Harris<jstevh@msn.com><3c65f87.0403140709.69290995@
posting.google.com>:> Unfortunately Dik Winter seems intent on
continuing to use my writing> on his webpages without my
permission.> My guess is that he feels that there's little I
can do about it from> here as he's in the Netherlands.> I need
advice on handling this person.> How do you get someone to obey
basic laws when they're hiding out in> another country?> Here's
one of his webpages:>
http://homepages.cwi.nl/~dik/english/mathematics/jsh.html> And
no matter how much you hate me, I hope that you understand
that> there's a reason why people aren't allowed to just copy
someone else's> writing and publish it at their whim.> And
part of it is common decency.> What can I do about Dik
Winter?> James HarrisAsk him politely to remove it, via
certified mail that he cansign and return. If he refuses, get
a lawyer and sue.-- #191, ewill3@earthlink.netIt's still legal
to go .sigless.===
===
Subject:
: Re: Copyright Issue: Dik Winter
publishing my work> In sci.math.num-analysis, James Harris>
<jstevh@msn.com
<3c65f87.0403140709.69290995@posting.google.com>:>
Unfortunately Dik Winter seems intent on continuing to use my
writing> on his webpages without my permission. My guess is
that he feels that there's little I can do about it from> here
as he's in the Netherlands. I need advice on handling this
person. How do you get someone to obey basic laws when they're
hiding out in> another country? Here's one of his webpages:
http://homepages.cwi.nl/~dik/english/mathematics/jsh.html And
no matter how much you hate me, I hope that you understand
that> there's a reason why people aren't allowed to just copy
someone else's> writing and publish it at their whim. And part
of it is common decency. What can I do about Dik Winter?> James
Harris> Ask him politely to remove it, via certified mail that
he can> sign and return. If he refuses, get a lawyer and sue.>
-- > #191, ewill3@earthlink.net> It's still legal to go
.sigless.James doesn't know how to do anything politely,
however, he expects everyoneelse to treat him politely.David
Moran===
===
Subject:
: Re: Copyright Issue: Dik Winter publishing my
workhere I am, back in the juvie hall of mathematics. anyway,
someone surely already said this, butit looks as if Dik just
did some original, andperhaps more concise, work, citing
yours, butnot copying it verbatim. (last that I
heard,equations are uncopyrightable, at least ifthey're not
attached to some algorithmic application; or,in other word,
Decategorified !-) your lawyer is going to need something
else, orto be an awfully talented specialist in copyright
--and *we* have the anti-SLAPP law, at least in California. >
What can I do about Dik Winter? > Ask him politely to remove
it, via certified mail that he can> sign and return. If he
refuses, get a lawyer and sue. --Give Earth a Trickier Dick
Cheeny -- out of office, after GIGA
years.http://www.benfranklinbooks.com/http://www.rand.org/
publications/randreview/issues/rr.12.00/http://
members.tripod.com/~american_almanac===
===
Subject:
: Re: Copyright
Issue: Dik Winter publishing my work> Unfortunately Dik Winter
seems intent on continuing to use my writing> on his webpages
without my permission.My guess is that he feels that there's
little I can do about it from> here as he's in the
Netherlands.You could ask your president to invade the
Netherlands.Of course you will probably need to show him the
country on a map.Marc Olschok===
===
Subject:
: Re: Copyright Issue:
Dik Winter publishing my work > Unfortunately Dik Winter seems
intent on continuing to use my writing > on his webpages
without my permission.... > Here's one of his webpages: >
http://homepages.cwi.nl/~dik/english/mathematics/jsh.html >
And no matter how much you hate me, I hope that you understand
that > there's a reason why people aren't allowed to just copy
someone else's > writing and publish it at their whim.I note
that that page is *not* a copy of a page by you, but a
paraphrase(i.e. modified to show with your reasoning that x^5
+ y^5 = n.z^5 isfalse). On jsh1.html you will find extensive
explanation of *why* it wascritique on jsh3.html. jsh4.html
contains again a paraphrase of somethingInteresting though
that since you posted your complaint the number ofhits has
increased... (But it still has a measly 133 hits and is nowat
15th place. In February it was not even in the top 25 of my
pages.)-- dik t. winter, cwi, kruislaan 413, 1098 sj
amsterdam, nederland, +31205924131home: bovenover 215, 1025 jn
amsterdam, nederland; http://www.cwi.nl/~dik/===
===
Subject:
: Re:
Copyright Issue: Dik Winter publishing my work<... 2. ... His
use is non-profit, and it will have> no negative effect on the
potential market value of your work (the market> for elementary
mathematics errors obfuscated by non-standard notation and>
terminology is rather small for some reason).There's the humor
magazine market to consider.===
===
Subject:
: Re: 5 circlesI should
have followed the description of the constructionin a past
issue of, I think, *21st CCE Science and Tech.*,as I had meant
to return to it.> to construct the pentagramma mirificum.> if
the i'th great circle interscts the (i-1) and (i+1) circles>
at 90 degrees, what is the angle of intersection> of those
two? Usama bin Laden is in a perfectly undead
superposition,with reference to the Bush and Media Hype,
althoughit wouldn't matter weather this state collapsedto dead
or alive, to Blair's McCrusade ... in the venueof
Schroedinger's joke about the Cat in the Boxwith the Atom to
Its Head .-) thus quoth:just doesn't make sense., even to the
phyicists! It seems that as we approach the limits of our
current understanding of themicro and macrocosms, many of our
concepts begin to break down. Buckycontended unflinchingly
that the problem was our use of Cartesian --Give Earth a
Trickier Dick Cheeny -- out of office, after GIGA
years.http://www.benfranklinbooks.com/http://www.rand.org/
publications/randreview/issues/rr.12.00/http://
members.tripod.com/~american_almanac