mm-57
===
An efficient way to compute the variance uses the
raw-score formula of statistics. If you take the defn. of
variance formula and expand it out and replace mu with its
estimate (sample mean) you get the raw-score formula in terms
of computation, you will only require the sum of the sample
and thesum of the squares of the sample along with the number
of samples.Clay> Dear all, In one of my program, I need to
compute the variance of a 8x8 data matrixas> fast as
possible... Any fast algorithm with lowest complexity? By the
way, since what I actually need is an indication of signal
local> activity, is there any other measure which can also be
a indicator o? activity and is less complex than
computing variance?> -Walala
===
 In one of my program, I
need to compute> the variance of a 8x8 data matrix as> fast
as possible...What does it means ?You just have 16 samples of
a signal and you want the variance ?For me it is not clear why
deviation fromthe mean is a good indicator of
activity.Obviously, there is no fast algorithm for computing
a sumof 16 terms !BUT if you compute the variance on a moving
windows, thereis a very fast algorithm, that you can ?ure out
pretty easily.Your just have to perform :1. compute mean m and
replace your image I by M=(I-m)^22. compute moving average
along X direction. This can be doneby a recursive algorithm,
because if S(k) is the moving averageand F(k) is the initial
function (e.g. a row of M), you haveS(k) = S(k-1)+F(k)-F(k-9)
(here I assume window of size 8)3. compute moving average
along Y direction.All this stuff needs about O(n) operations
where n is the number of pixelsin the original image and it's
independant of the size of the window.Gabriel
===
> You just
have 16 samples of a signal and you want the variance ?Sorry,
I meant 64 samples.> Obviously, there is no fast algorithm for
computing a sum> of 16 terms !Sorrrry once again 8x8=64 not 16
...G.
===
We look at
the usual long multiplication of, for example, 123*48. 123 x
48 ------- 984 +492------- 5904. However, I wish to
concentrate just on the tens and units ?ures of the
resulting product. This is called ?multiplying modulo 100.'
That is, any whole number of 100s can be thrown away in the
process. Then, (100+23)*48 == 48*100+23*48 (is congruent
to)== (20+3)*(40+8) == 2*4*100+3*40+8*20+3*8 == 120+160+24 ==
104 == 04 modulo 100. In calculating the remainder of product
(integers w*z) after dividing by positive integer, t, it can
often be made easier by taking away convenient multiples of
the modulus, t, from w or z respectively before executing the
multiplication. Let w = at+b, say, and z = ct+d, say. Then, wz
= (at+b)(ct+d) = at(ct+d) +b(ct+d) = t(act +ad +bc) +bd == bd
modulo t. Bd should usually be less than w*z, but it may
require to be reduced further. I photographed car number
plate ?MP163' at Hanson St, Newtown, Wellington on today,
26-5-2002. This number plate could be shorthand for Mersenne
numbers, 2^prime-1. (Figure 2 raised to a prime index,
subtract 1.) The world's largest known prime number in 2001
often is2^13.4million-1. (Greater than 13 million 2s
(Dominion Post, NZ Herald, GIMPS greatinternet mersenne prime
search.)By the way, the 40th known Mersenne prime exponent isa
twin prime, 10x(23x63)^2 +/-1. And a previous recordholder was
virtually a palindrome, Table Mountain,
increasingdecreasing.Exponent (1*2*34*44432+1)= 3021377.
[sic.] _____ / A theorem of mathematics number theory says
if(a is a positive integer )..a^prime-1 has factor/s they
must be of the form 2*k*p+1. However, the Penguin Dictionary
of Curious and Interesting Numbers (1997), entry 28, table of
perfect numbers, does not give M_163 as a Mersenne prime.
Therefore, I have written myprogram ?powabc1' in BBC
interpreted Basic64, which tests for just such factors.(On
Acorn A5000 computer, UK 1990, RISC OS-reduced instructionset
computer, 4 Megabytes RAM randomaccess memory.)In this case,
several programs found 150287 and 704161, etc. divide M_163.
Or 2^163 == 1 modulo 150287. Have I found a factor of M_163?
The following, I believe, shows that I have.Check. 2^163=
2^3*(2^20)^8 (is congruent to)==
8*1048576^82^20==1048576-6*150287 == 146854 mod 150287 ==
-3433. Squaring 2^40= 11785489== 63103 Squaring
2^80==3,981,988,609 ==134544== -15743 Squaring
2^160==247,842,049 ==18786 2^163==8*18786 ==150288==1 mod
150287. Q.e.d.By the way, my methods found (the least prime)
factor1580,187,223 of (10^9)^(10^9)+3. (sci.math sensation
2001.)This is, digit 1 followed by 9 billion zeroes.
Surpassedonly by Graham's number. I suspect gigaplex
=10^billion. (There is a typo error in Penguin Dictionaryof
Curious and Interesting Numbers, 1997.) yours sincerely, /
Donald S. McDonald (Wellington, New Zealand)
===
There was a
time when people spoke their minds, and were not afraid to
offend -and that since then, too many truths have been
buried.Mark Kurlansky, 1968: The Year That Rocked The
World(Ballantine, 2004) thanks to Frank LauriaIn 1968 I was
in La Jolla CA at UCSD during time Greg Benford describes
inTimescape living on Bonair Street Wind an Sea Beach near
UnicornTheater with Ken Kesey's Merry Prankster Bus parked
nearby frequently.I was also teaching at San Diego State with
Fred Alan Wolf....PZ: I can't imagine anything more
wrong-headed. And you say Rovelli is a bigshot?That is why I
say Rovelli's position is incoherent.JS: Is coherence in the
mind of the beholder?PZ: If you want to treat g_uv as a
physical ?ld, based on its dynamical character(i.e.
matter-dependence), then the natural thing to do is separate
the backgroundgeneralized Minkowski kinematical metric (which
is NOT matter dependent) aschronogeometric, and treat the
gravitational g_uv alone as physical -- byRovelli's own
argument.JS: That will not work. It will not tip the light
cones. It will not give the correct bending of light.Also you
are too vague on what you mean by matter on the RHS
ofGuv(Einstein) = -(alpha)(alpha')Tuv(Matter)alpha = e^2/hc ~
1/137alpha' = 8pi(Witten's reciprocal string 
tension)alpha' ~
(10^-32 cm)^2 in a common convention.I also include on
RHStuv(Vacuum) ~ [(alpha)(alpha')]^-1/zpfguv/zpf =
Lp^-1[Lp^3|Vacuum Coherence|^2 - 1]Repulsive dark energy is
/zpf > 0Attractive dark matter is /zpf < 0FRW Omega(Dark
Energy + Dark Matter) ~ 0.96With Omega(Total) = 1 i.e. FLAT
SPACEPreferred foliation is where CMB is maximally isotropic
to ~ 10^-5.This allows accurate navigation with weightless
warp drive and traversable wormholes both supported by
con?urations of dark energy and dark matter exotic w = -1
vacua.PZ: Yet he doesn't even mention this.JS: There was a
time when people spoke their minds, and were not afraid to
offend -and that since then, too many truths have been
buried.Mark Kurlansky, 1968: The Year That Rocked The
World...JS: Also at quantum level Rovellimention's 
Dirac's
insight that the Heisenberg Picture is better thanSchrodinger
Picture and that time as in the ? time's arrowis not
dynamical time but is a statistical thermodynamic
construct.PZ: Sure, if you want to quantize everything and
abandon time.Unruh's for example, and some of the
others.Unruh has some important ideas however.PZ: But then I
fail to see any material distinction between relabelingthe
bare unindividuatedspacetime points (passive diffeomorphism),
on the one hand, andshifting all physical?lds (including the
GR metric ?ld) with respect to such a rawmanifold
(activediffeomorphism), on the other. Kretschman's point
appears to be fullyvalid in both cases:how can physics in
general -- any sensible physics -- possibly dependon a mere
re-labelingof raw unindividuated spacetime points; or, for
that matter, on acommon shift of all physicalsystems,
including the *physical* metric ?ld g_uv, with respect
tosuch a set of unindividuatedpoints?JS: Admittedly a sticky
wicket that I also need to understand moredeeply.PZ: It's an
arti?ial model that has nothing to do with physical
relativity IMO.JS: I found this remark by Goldstein
helpful:In the ADM formulation 4-diffeomorphism invariance
amounts to the requirement that one ends up with the same
space-time, up to coordinate transformations, regardless of
which path in multi-?gered space-time is followed, i.e.
which lapse function N is uses. p.278OK so this idea goes
back to the archetypal notions of classical thermodynamics
with a state function, to holonomic integrability of equality
of mixed partial derivatives with not multiply-connected
manifolds, to a closed Cartan exterior differential form on a
cycle (no boundary), no topological defects and all that.The
passive coordinate transformations are like EM gauge
transformation (e/c)Au -> (e/c)Au + hChi,u in a ?ed gauge
constraint.PZ: Rotating a system in isotropic space is
connected with a true physical symmetryof the system
including the vacuum in which it is embedded. Yes, the
symmetryof the system Hamiltonian under such a transformation
is an active transformationalsymmetry that is characteristic
of the particular system -- unlike the invariance of
theproper formal expression of any physical law under a
*mere* coordinate transformation.JS: Note also 12.2.2 p.279
alluding to your digging up Kretchmann from Dr.
Frankenstein's favorite graveyard. ;-)The fundamental
symmetry at the heart of general relativity is its invariance
under general coordinate transformations of spacetime. It is
important to stress that almost any theory can be formulated
in such a 4-diffeomorphism invariant manner by adding further
structure to the theory (e.g. a preferred foliation of
spacetime as a dynamical object). General relativity has what
is sometimes called serious diffeomorphism invariance, meaning
that it involves no spacetime structure beyond the 4-metric
and, in particular, singles out no special foliation of
spacetime.Goldstein and Teufel then knock standard QGrav
including even Ashtekar -> Loop Spin Foams that are perhaps
near to being falsi?d by NASA's EINSTEIN inJack better check
this story out:Gary S. Bekkum...How to getlocalization in
space and the ? time as we experience it in ourimmediate
inner consciousnesshas nothing to do with the particular local
coordinate representationlike r and t in, for example,K =
e^2GM/c^2rdr/dt = c/Kfor null geodesicand in his Tables
generally in the context of potentially practicalmetric
engineering of the guv ?ld using the EM Au ?ld in spite
ofthe enormous gravity string tension ~ c^4/G ~ 10^19 Gev per
10^-33 cm.PZ: You are blocking his actual de?ition of r.
Classic operationalism does notapply to a theory of this
type. That is a critical point. Miss that and of
coursenothing makes sense.JS: I am not ready to renounce PW
Bridgman's Operationalism. Indeed, nothing Hal Puthoff says
about the foundations of his PV makes any sense to my mind.
If it ain't broke, don't ? it. Of course I am not a
doctrinaire positivist like Stephen Hawking proudly proclaims
he is....JS: Correct, with the proviso that matter includes
both real, i.e.on mass shell, sources as well as virtual,
i.e. off mass shellsources. The virtual sources divide into
two classes:I. Non-exotic near EM ?ld Fuv giant coherent
quantum states ofvirtual photons that contribute to
Omega(Matter) of the FRW metric andto Tuv in Einstein's local
?ld equation.II. Exotic vacuum w = -1 zero point
stress-energy density local tensor~ (String Tension)/zpfguv
for both repulsive dark energy /zpf > 0 ofnegative pressure
and attractive dark matter /zpf < 0 of positivepressure.These
exotic vacuum virtual sources contribute Omega(Exotic Vacua)
~0.96 to Omega(Total) = 1 in our large-scale spatially
?t-in?ary local Level I Hubble sphere brane world
as inLenny Susskind's megalopolis Landscape subject to the
naturalselection of the Weak Anthropic Principle (WAP)OK, I
think I made an error above including brane worlds in sense
of parallel worlds?Here is why I think I made an error (If I
did so did Hawking and Scienti? American in their popular
science reports):D-Branes are extended surfaces without
edges. In order that the black hole be a localized object, it
is assumed that our ordinary four dimensions (three space and
one time) are all orthogonal to these D-Brane surfaces ...
Thus to us these D-Branes would look as though they were
located at a point (or at least a very small region) of our
observable three dimensions of space. W. G. Unruh p. 168
Black holes, dumb holes, and entropy, i.e. the D-Branes are
in the compacti?d Calabi-Yau space. I have to look again at
Hawking's The Universe in a Nutshell that seems to give the
wrong idea here? Perhaps I mis-remember?Note also Ed Witten's
formula generalizing Heisenberg's quantum uncertainty
principle, i.e. eq. (5.9) p. 136Delta X > h/DeltaP +
alpha'(DeltaP)/hThe second gravity-string source of
uncertainty should give the irreversible statistical arrow of
time not found when alpha' = 0 i.e. in?ite string tension, or
in?ite space-time stiffness of action without reaction as is
also found in the signal locality of orthodox quantum theory
in sense of Antony Valentini's papers.Mass without mass, but
with a strong micro-gravityG* ~ 10^40G on 1 fermi scale. The
wormhole has an attractive dark matter exotic vacuum core
whereVacuum Coherence --> 0just like inside a quantized
vortex of circulation in super?eII.Therefore, in the
core/zpf ~ - (1 fermi)^-2w = -1Therefore the quantum pressure
is positive and the exotic vacuum core gravitates
asGrad^2V(Exotic Vacuum) ~ -c^2(1 fermi)^-2this prevents the
spread out electric charge from exploding and also
compensates any quantized rotation centrifugal forces. For
now keep charge and rotation zero for simplicity. That is I
here only model a spin 0 neutral micro-geon. That means I
would need a high-power graviton laser as the Heisenberg
uncertainty scattering probe microscope. That's OK since this
is only a gedankenexperiment.The SSS metric then has the
factor1 - 2GM/c2rWhere G(mass density) is replaced by
c^2/zpfGM is replaced by ~ c^2/zpfR^3for a micro-geon of size
Rr is DeltaX as in Witten's formula above with DeltaP as the
scattering momentum transfer between gravitons and the
micro-geon.Therefore, the Schwarzschild factor is (neglecting
factors of 2, pi etc all lumped into dimensionless parameter
b1 - bc^2/zpfR^3/DeltaXThe critical value of Delta X is then
the event horizon where the Schwarzschild factor vanishes -
can it be reached?This micro-geon is a solution of the exotic
vacuum local ?ld equationGuv(Einstein) + /zpfguv = 0At
critical Delta X, the geon looks like a POINT PARTICLE from
the huge space-warp induced by the probe's momentum transfer
Delta P.C is the circumferencedC/dR = 2pi (1 -
bc^2/zpfR^3/DeltaX)^1/2Where one considers the radial size
?ed at R the scale of the throat of the wormhole.So when can
we haveDeltaX = bc^2/zpfR^3 ?Note that alpha' may be large of
order (10^-11 cm)^2 not (10^-32 cm)^2 as in Witten's
idea.
===
Interesting!Jack,Superstring theory does not view
the Planck scale of spacetime as a quantumfoam but rather as
the Planck scale is approached the dimensionality ofspacetime
goes to 10-d. In string theory the spacetime does ?te
butthis ?tion is harmonic rather than chaotic. In fact
the assumptionthat the harmonics of the strings absorbs all
the quantum ?tion can beused to derive the
dimensionality of spacetime. Also Lorentz invariance
isassumed in this calculation. This was ?st done by L. Brink
and H.B.Nielsen in 1973 (A Simple Physical Interpretation of
the Critical Dimensionof Space-time in Dual Models, *Physics
Letters* 45B:4 (1973) 332-336. Thispaper is also included in
the anthology edited by John Schwartz,*Superstrings: the
First 15 Years of Superstring Theory, Vol. 1*
(WorldScienti?, 1985). [Note that Dual Models is the old
name for stringtheory, when it was still evolving away from
the terminology of s-matrixtheory. However, the Mandelstam
labels of S, T, and U duality haveresurfaced in membrane
theory!] I mentioned the Brink-Nielsen view of string theory
in my paper Notes onpdf.] Also on page 6 of F.W. Stecker's
pdf paper [referred to in the NASAreport], he says We note
that there are variants of quantum gravity andlarge extra
dimension models which do not violate Lorentz invariance and
forwhich the constraints considered here do not apply. BTW:
In the Acknowledgments on page 8, Serge Rudaz is one of
fourpeople thanked for helpful discussions. I remember
meeting Serge Rudaz inseminars. He was then a young physics
student (at Cornell?) who was an oldacquaintance of yours. He
told us about instantons -- which was a new ideathen. I have
not heard of him since 1976 -- until seeing
thisacknowledgement! Nuff said! Saul-PaulJS: Yes, Serge was
with me at UCSC in Summer 1973 when I went to see Jean
Cocteau's Orphee on campus with Helen Quinn (who was close to
having he baby at that time) and I think Serge and a few
others. This was days before I went to SRI to meet with
Puthoff and Targ on the tape you have. The story is in the
book Destiny Matrix.----------Therefore quantum foam is
suspect. Also Bohm's quantum potentialview of vacuum
?tions is relevant in context of the recentpaper from
Teheran.I need to follow the experiment below more
carefully.Jack better check this story out:Gary S.
Bekkum
===
i have realised a few mathematical functions?,
properties?, lately andwas wondering about them, such as
whether they are already known, orare even important.if
anyone is interested, hit me up,AIM:heartxenocide
===
I really
hope no one takes this as spam, I'm a student member of
theMAA, and I've posted here a few, brief times, however
since I'm stilla lower division undergraduate most of the
subject matter is far abovemy head, so I'm not a regular
(yet!)With that said, I have created some math t-shirt
designs that justmake me smile and I'd like to share them
with the math community, withthe hopes that they delight
someone else as much as they delight me.the url is
http://www.cafeshops.com/subjective and the designs areunder
Math and Science (click on the Einstein picture)All comments,
critiques, commendations, cat-calls and creative input
===
I am
stuck in high school maths mode, and can't seem to get into
university level maths. This might be because I am entirely
self taught, but I don't know. Does anyone else have this
problem?I am at a level where I understand most high school
maths, and I've studied Calculus Made Easy. It would be
helpful to have some kind of way to check my knowledge.Does
anyone know of any good textbooks that cover high school
maths with worked exercises, and any texts that help the
transition from high school maths to the more exciting stuff
at university level? A book that takes time to explain
things, point out applications and say why rather than just
how. I am trying to understand maths, not just learn some
techniques or shortcuts.By the way, I would prefer internet
resources and small paperback books. There's no way I can
afford college textbooks unfortunately.Johnathan
===
 I am
stuck in high school maths mode, and can't seem to get into>
university level maths. This might be because I am entirely
self> taught, but I don't know. Does anyone else have this
problem? I am at a level where I understand most high school
maths, and I've> studied Calculus Made Easy. It would be
helpful to have some kind of> way to check my knowledge. Does
anyone know of any good textbooks that cover high school
maths> with worked exercises, and any texts that help the
transition from> high school maths to the more exciting stuff
at university level? A> book that takes time to explain
things, point out applications and say> why rather than just
how. I am trying to understand maths, not just> learn some
techniques or shortcuts. By the way, I would prefer internet
resources and small paperback> books. There's no way I can
afford college textbooks unfortunately. JohnathanYou could
check
out:http://store.doverpublications.com/
by-subject-science-and-mathematics-mathematics-calculus.htmlth
ey are cheap and useful. You could also try Schaum's
outlines; they arealso very good.For web content,
try:http://www.math.temple.edu/~cow/If you run a google
search, you will ?d a plethora of math relatedwebsites.
Many, many, many more than I care to list. Make google your
bestfriend. You can ?d anything you could possibly want to
know using googlein about 30 seconds.Lurch
===
> I am stuck
in high school maths mode, and can't seem to get into>
university level maths. This might be because I am entirely
self> taught, but I don't know. Does anyone else have this
problem?> I am at a level where I understand most high
school maths, and I've> studied Calculus Made Easy. It would
be helpful to have some kind of> way to check my knowledge.>
> Does anyone know of any good textbooks that cover high
school maths> with worked exercises, and any texts that help
the transition from> high school maths to the more exciting
stuff at university level? A> book that takes time to explain
things, point out applications and say> why rather than just
how. I am trying to understand maths, not just> learn some
techniques or shortcuts.> By the way, I would prefer
internet resources and small paperback> books. There's no way
I can afford college textbooks unfortunately.> JohnathanYou
could check out:>
http://store.doverpublications.com/
by-subject-science-and-mathematics-mathematics-calculus.htmlth
ey are cheap and useful. You could also try Schaum's
outlines; they are> also very good.For web content,
try:http://www.math.temple.edu/~cow/If you run a google
search, you will ?d a plethora of math related> websites.
Many, many, many more than I care to list. Make google your
best> friend. You can ?d anything you could possibly want to
know using google> in about 30 seconds.LurchIn particular,
include ext:pdf site:.edu in your searches. You'lljust get
postscript notes, usually written up by
professors.'cid
===
There was a time when people spoke their
minds, and were not afraid to offend -and that since then,
too many truths have been buried.Mark Kurlansky, 1968: The
Year That Rocked The World(Ballantine, 2004) thanks to Frank
LauriaIn 1968 was in La Jolla CA at UCSD during time Greg
Benford describes inTimescape living on Bonair Street Wind an
Sea Beach near UnicornTheater with Ken Kesey's Merry Prankster
Bus parked nearby frequently.I was also teaching at San Diego
State with Fred Alan Wolf.OK, now I have looked at the rest
of Rovelli's argument, I'll work throughthis one more
time.PZ: Rovelli's position is very odd. First he says:...Of
course, nothing [in GR] prevents us... from singling out
thegravitational?ld as ?the more equal among equals', and
declaring that location isabsolute inGR, because it can be
de?ed with respect to it. (p 108)JS: But he rejects that
Paul. He says doing that misses the greatEinsteinian
insight.The great Einsteinian insight being the conditional
and physical nature ofthe gravitational metric ?ld -- which
con?with the great Einsteinian insightstrict
equivalence.The opposite of a profound truth is another
profound truth. -- Niels BohrWhat Rovelli doesn't seem to
understand is that this all makes perfect sense onceyou give
up strict equivalence and distinguish the background and
physical metrics.JS: I do not understand this distinction.
Please give more details what you mean.Have you read pp. 112
- 114 that completely demolishes Hal Puthoff' suse ofdr/dt =
c' = c/K radial null geodesicin his Tables.PZ: It does no
such thing. I would not even characterize pp 112-114 as an
argument.It is simply a sketch of a model in which
*everything* is quantized except the rawmanifold.JS: It shows
no intrinsic meaning to Puthoff's r and t as he means it in
his Tables.PZ: He wants to throw away time in order to keep a
uni?d g_uv.Why do you think this is an argument against c' =
c/K? As far as I can see it is simplya different
theory.Rovelli brie?siders an alternative approach in
which we retain the Minkowskibackground of standard QFT. But
there he hasg_uv = n_uv + ?tionswhich makes no sense to
me. He does not seem to understand the distinction
betweenkinematical g_uv and dynamic gravitational g_uv. He
cannot get his head over the uni?dmetric.What does he mean
by ?tions?JS: What do you mean by kinematical g_uv and
dynamic gravitational g_uv apart from Ruvwl = 0 in the former
and not in the latter.JZ: Sounds like sheer nonsense to me.
Nonsense because it is divorced from the
physicalfundamentals.JS: No argument from me on that one.PZ:
This of course is exactly what the classic
Einsteinchronogeometric model does,going all the way back to
special relativity.JS: I think you are misreading Rovelli.PZ:
The Einsteinian model is a chronogeometric model, in which the
metric g_uv re?thefundamental structure of spacetime.JS:
If you mean, for example,dT = goo^1/2dtdR = grr^1/2drThen I
agree that dT and dR are physical and dt and dr are
not.However, to get a gravity shift of light frequency
dodT'/goo'^1/2 = dT/goo^1'2treating dt as a 
kind of nonlocal
invariant.If you mean more than this, then explain with
detailed examples.PZ: That is the great Einsteinian insight
-- which is.unfortunately, based on strict Einstein
equivalence, which is ?titious.JS: Again I really do not
understand what you mean by this sentence.PZ: Now Rovelli
wants to pretend that the great Einsteinian insight is
something else entirely-- that the gravitational ?ld is a
physical ?ld.Loony tunes. He is tying himself up in
knots.JS: Yes, in global specialrelativity, NO in local
general relativity.PZ: In Einstein general relativity, the
uni?d metric g_uv represents the fundamental structureof
spacetime, and in inseparable from it. That's Einstein.It
doesn't even make sense to me to say that this
chronogeometric model holds only locally.PZ: That is
precisely what distinguishes theEinsteinian from the
Lorentzian model. The transformational andmetric structureis
not, in the classic Einsteinian model, separable from
spacetimeitself.But then he says:There is no absolute
referent of motion in GR; the dynamical ?ldsmove withrespect
to each other. (p 108)JS: Again you are misreading. There is
no contradiction here inRovelli's argument.PZ: This is like
arguing that everything is relative because everything is
relative to theabsolute, including the absolute.Sounds like
Lewis Carroll.JS: ;-)PZ: None of this has anything to do with
Einstein general relativity, which treats theinertial ?ld as
real. The uni?d metric derives its physical meaning from
thisequivalence.Rovelli ignores all this and simply takes the
uni?d Einstein g_uv as given.PZ: This second assertion seems
to approach the metric ?ld of GR asjust another ?ld,
whichis very close to and even indistinguishable from the
physical rubberrod and clock modelof PV and Yilmaz -- just
another physical ?ld which happens to havemetric
properties.And of course such a ?ld is fully relational with
respect tounindividuated spacetime pointson a raw manifold
stripped of all coordinate systems, transformationproperties,
and metrics.JS: Yes on just another ?ld. But NO that it's
like PV and Yilmaz. Nottrue at all because,at least in PV,
Hal uses an absolute non-dynamical background globalMinkowski
spacePZ: That is what Rovelli *should* be doing, but he
doesn't even consider thispossibility. He seems to think you
can treat uni?d g_uv as a physical ?ld.JS: Why do you think
you cannot?PZ: I can't imagine anything more wrong-headed. And
you say Rovelli is a bigshot?That is why I say Rovelli's
position is incoherent.JS: Is coherence in the mind of the
beholder?PZ: If you want to treat g_uv as a physical ?ld,
based on its dynamical character(i.e. matter-dependence),
then the natural thing to do is separate the
backgroundgeneralized Minkowski kinematical metric (which is
NOT matter dependent) aschronogeometric, and treat the
gravitational g_uv alone as physical -- byRovelli's own
argument.JS: That will not work. It will not tip the light
cones. It will not give the correct bending of light.Also you
are too vague on what you mean by matter on the RHS
ofGuv(Einstein) = -(alpha)(alpha')Tuv(Matter)alpha = e^2/hc ~
1/137alpha' = 8piWitten's reciprocal string 
tension)alpha' ~
(10^-32 cm)^2 in a common convention.I also include on
RHStuv(Vacuum) ~ [(alpha)(alpha')]^-1/zpfguv/zpf =
Lp^-1[Lp^3|Vacuum Coherence|^2 - 1]Repulsive dark energy is
/zpf > 0Attractive dark matter is /zpf < 0FRW Omega(Dark
Energy + Dark Matter) ~ 0.96With Omega(Total) = 1 i.e. FLAT
SPACEPreferred foliation is where CMB is maximally isotropic
to ~ 10^-5.This allows accurate navigation with weightless
warp drive and traversable wormholes both supported by
con?urations of dark energy and dark matter exotic w = -1
vacua.PZ: Yet he doesn't even mention this.JS: There was a
time when people spoke their minds, and were not afraid to
offend -and that since then, too many truths have been
buried.Mark Kurlansky, 1968: The Year That Rocked The
WorldPZ: So he has not made any argument at all against a
bimetric approach. He has simplyignored it, even though his
own argument points to it.JS: with a literal meaning for
coordinates r & t, otherwise his formuladr/dt = c/Kis
physically meaningless, which it is in GR as explained by
Rovelli onpp. 112 - 114.PZ: Only if you insist exclusively on
operational meaning -- which would also knock outF = ma, which
by a similar argument is empirically meaningless.JS: Yes,
until you also add things likeF = - GMmr/r^3F = eE +
e(v/c)xBetc.That is, Hal's use of engineering is precisely
missing the thirdstep on Rovelli's p. 108.PZ: 
Rovelli's third
step is merely a sketch of a proposal to quantize everything,
whileretaining a uni?d g_uv treated in its entirety as a
physical ?ld.This is not an argument.Rovelli's argument for
his proposal, such as it is, is to be found on p 109:In my
*opinion*, this is the right way to go.Well, in my *opinion*,
it's fundamentally wrong-headed. In addition, Rovelli
hasgrossly mischaracterized the Feynman-type alternative --
which shows that he hasnot understood it; although he is
right that under this alternative, the GR
conceptualrevolution is indeed dis-valued, but to a much
greater extent than he seems to realize.JS: Also p. 112Recall
that Einstein described his great intellectual struggle to
?dGR as ?understanding the meaning of coordinates'.This is
my key objection to Puthoff's and Ibison's use of r 
ant tin
their PV modeling.They do not IMHO understand the meaning of
coordinates: in GR. Neitherdoes Yilmaz apparently.PZ: I have
no idea of what you are getting at here. Einstein's point
about coordinatesdepends entirely on his strict equivalence
thesis, which is untenable.JS: I guess I have never
understood your idea here. Can you try to explain it again
withcomplete clarity?PZ: Once you abandon strict equivalence,
the uni?ation of gravitational and inertialg_uv is formal and
arbitrary. It has no deep physical meaning. Then we
recoverchronogeometric background coordinates together with a
physical rubber-rod-and-clock non-inertial g_uv.Rovelli
doesn't seem to understand that.JS: Neither do I.pp 112 - 114
make this clear IMHO.PZ: This is not an argument -- it's
merely a proposal: In my opinion....JS: Also at quantum level
Rovellimention's Dirac's insight that the Heisenberg 
Picture
is better thanSchrodinger Picture and that time as in the ?ti
me's arrowis not dynamical time but is a statistical
thermodynamic construct.PZ: Sure, if you want to quantize
everything and abandon time.Unruh's for example, and some of
the others.Unruh has some important ideas however.PZ: But
then I fail to see any material distinction between
relabelingthe bare unindividuatedspacetime points (passive
diffeomorphism), on the one hand, andshifting all
physical?lds (including the GR metric ?ld) with respect to
such a rawmanifold (activediffeomorphism), on the other.
Kretschman's point appears to be fullyvalid in both cases:how
can physics in general -- any sensible physics -- possibly
dependon a mere re-labelingof raw unindividuated spacetime
points; or, for that matter, on acommon shift of all
physicalsystems, including the *physical* metric ?ld g_uv,
with respect tosuch a set of unindividuatedpoints?JS:
Admittedly a sticky wicket that I also need to understand
moredeeply.PZ: It's an arti?ial model that has nothing to do
with physical relativity IMO.JS: I found this remark by
Goldstein helpful:In the ADM formulation 4-diffeomorphism
invariance amounts to the requirement that one ends up with
the same space-time, up to coordinate transformations,
regardless of which path in multi-?gered space-time is
followed, i.e. which lapse function N is uses. p.278OK so
this idea goes back to the archetypal notions of classical
thermodynamics with a state function, to holonomic
integrability of equality of mixed partial derivatives with
not multiply-connected manifolds, to a closed Cartan exterior
differential form on a cycle (no boundary), no topological
defects and all that.The passive coordinate transformations
are like EM gauge transformation (e/c)Au -> (e/c)Au + hChi,u
in a ?ed gauge constraint.PZ: Rotating a system in isotropic
space is connected with a true physical symmetryof the system
including the vacuum in which it is embedded. Yes, the
symmetryof the system Hamiltonian under such a transformation
is an active transformationalsymmetry that is characteristic
of the particular system -- unlike the invariance of
theproper formal expression of any physical law under a
*mere* coordinate transformation.JS: Note also 12.2.2 p.279
alluding to your digging up Kretchmann from Dr.
Frankenstein's favorite graveyard. ;-)The fundamental
symmetry at the heart of general relativity is its invariance
under general coordinate transformations of spacetime. It is
important to stress that almost any theory can be formulated
in such a 4-diffeomorphism invariant manner by adding further
structure to the theory (e.g. a preferred foliation of
spacetime as a dynamical object). General relativity has what
is sometimes called serious diffeomorphism invariance, meaning
that it involves no spacetime structure beyond the 4-metric
and, in particular, singles out no special foliation of
spacetime.Goldstein and Teufel then knock standard QGrav
including even Ashtekar -> Loop Spin Foams that are perhaps
near to being falsi?d by NASA's EINSTEIN inJack better check
this story out:Gary S. BekkumPZ: However, moving *everything
physical* -- including the uni?d g_uv -- along the
rawspacetime manifold is quite another kettle of ?h. It is
devoid of physical content IMHO.But again, a space-time frame
of reference is not *merely* a coordinate transformation,since
it represents the *motion* of a possible observer. Thus there
is no *a priori* reasonwhy physical laws should take the same
form in different frames of reference, contraryto Einsteinian
doctrine. Of course they *may* as a matter of fact, but that
is very differentfrom insisting a priori that they always
*should*.That's yet another Einsteinian red herring.PZ: What
is physically signi?ant here is not this abstract
andarti?ial de?ition of diffeomorphismwith respect to a raw
manifold conceived as an unindividuated pointset, but a shift
of physical?lds with respect to the metric-transformational
structure ofspacetime, which of course in GRdepends on the
distribution of matter.JS: Precisely, yes that is the idea I
think. It's a return to I thinkLeibniz's 
relationism?PZ:
Descartes. He is almost literally putting Descartes before
the horse ofphysical relativity.JS: ;-)How to getlocalization
in space and the ? time as we experience it in
ourimmediate inner consciousnesshas nothing to do with the
particular local coordinate representationlike r and t in,
for example,K = e^2GM/c^2rdr/dt = c/Kfor null geodesicand in
his Tables generally in the context of potentially
practicalmetric engineering of the guv ?ld using the EM Au
?ld in spite ofthe enormous gravity string tension ~ c^4/G ~
10^19 Gev per 10^-33 cm.PZ: You are blocking his actual
de?ition of r. Classic operationalism does notapply to a
theory of this type. That is a critical point. Miss that and
of coursenothing makes sense.JS: I am not ready to renounce
PW Bridgman's Operationalism. Indeed, nothing Hal Puthoff
says about the foundations of his PV makes any sense to my
mind. If it ain't broke, don't ? it. Of course I am 
not a
doctrinaire positivist like Stephen Hawking proudly proclaims
he is.PZ: So it appears that Rovelli ends up in a position
where he isessentially arguing that the matter-dependence of
the metric ?ld implies that the GR metric is really
aphysical metric, and the GRmetric ?ld is to be regarded as
a physical ?ld like any otherphysical ?ld.That is, the
*uni?d* metric ?ld. That is Rovelli's tacit, yet arbitrary,
assumptionIMO.Then, the physicsdepends only on the relative
disposition of the various physical?lds with respect to each
other.JS: Correct, with the proviso that matter includes both
real, i.e.on mass shell, sources as well as virtual, i.e. off
mass shellsources. The virtual sources divide into two
classes:I. Non-exotic near EM ?ld Fuv giant coherent quantum
states ofvirtual photons that contribute to Omega(Matter) of
the FRW metric andto Tuv in Einstein's local ?ld
equation.II. Exotic vacuum w = -1 zero point stress-energy
density local tensor~ (String Tension)/zpfguv for both
repulsive dark energy /zpf > 0 ofnegative pressure and
attractive dark matter /zpf < 0 of positivepressure.These
exotic vacuum virtual sources contribute Omega(Exotic Vacua)
~0.96 to Omega(Total) = 1 in our large-scale spatially
?t-in?ary local Level I Hubble sphere brane world
as inLenny Susskind's megalopolis Landscape subject to the
naturalselection of the Weak Anthropic Principle (WAP)OK, I
think I made an error above including brane worlds in sense
of parallel worlds?Here is why I think I made an error (If I
did so did Hawking and Scienti? American in their popular
science reports):D-Branes are extended surfaces without
edges. In order that the black hole be a localized object, it
is assumed that our ordinary four dimensions (three space and
one time) are all orthogonal to these D-Brane surfaces ...
Thus to us these D-Branes would look as though they were
located at a point (or at least a very small region) of our
observable three dimensions of space. W. G. Unruh p. 168
Black holes, dumb holes, and entropy, i.e. the D-Branes are
in the compacti?d Calabi-Yau space. I have to look again at
Hawking's The Universe in a Nutshell that seems to give the
wrong idea here? Perhaps I mis-remember?Note also Ed Witten's
formula generalizing Heisenberg's quantum uncertainty
principle, i.e. eq. (5.9) p. 136Delta X > h/DeltaP +
alpha'(DeltaP)/hThe second gravity-string source of
uncertainty should give the irreversible statistical arrow of
time not found when alpha' = 0 i.e. in?ite string tension, or
in?ite space-time stiffness of action without reaction as is
also found in the signal locality of orthodox quantum theory
in sense of Antony Valentini's papers.PZ: But while this may
allow or even imply a relational view of theraw spacetime
manifold, in theCartesian sense, it is clearly not physical
general relativity, inthe classic Einsteinian sense,which
requires fundamental identi?ation of the inertial
andgravitational metrics.JS: I do not understand what you
just said. EEP in every signi?antsense still holds IMHO.PZ:
You are in a loop.JS: Ground Hog Day. Help let me out of my
bottle Oh Thief of Baghdad. The Magic Flying Carpet is
obviously the weightless Alcubierre timelike geodesic faster
than light warp drive powered by dark energy and dark matter
exotic vacua con?urations. See my animated picture of this
in http://qedcorp.com/APS/StarGate1.movPZ: EEP is NOT
Einstein equivalence. EEP is merely a correspondence
principle.Einstein strict equivalence is NOT valid. And even
EEP is not strictly valid asadvertised (see e.g. Ohanian and
Ruf?i Ch. 1).given guv ?ld still works for example.PZ: This
simply re?weak equivalence -- equality of gravitational
and inertial mass.That does not imply Einstein equivalence.
It works whether we use a uni?d g_uvor go to a bimetric
formalism, and regardless of whether we consider inertial
forcesas real or ?titious.JS: The so-called ?titious or
inertial forces, e.g. G-Force when a jet takes off (0 to 120
mph in 2 sec) from an aircraft carrier are physically real. I
can tell you that from ?st-hand experience. The use of
?titious like the use of hidden variable are unfortunate
choices of words. Extra variable is better for the latter.PZ:
So this is not a valid argument IMO. It is an example of the
classical logical fallacyaf?ming the antecedent.JS: The
tidal stretch-squeezeworks etc.PZ: Of course. So?The point is
that you can always tell the difference between a real gravity
?ld andan inertial ?ld. They are NOT completely physically
equivalent.JS: If you mean Ruvwl =/= 0 locally for a real
gravity ?ld, and is zero for an inertial ?ld, I
agree.However, I do not see how you can writeguv =
guv(inertial) + guv(real)That is, apart fromguv = Globally
Flat Metric + du,v + dv,uI do not claim we can physically
distinguish the two terms on RHS nor is thesecond term small
compared to ?st.The Bohm pilot constraint for the extra
variable, i.e. actual distortion of Hagen Kleinert's
elastic-plastic 4D world crystal latticei.e. density of
vortex line topological defects of curvature disclination and
possibly torsion dislocations isdu = Lp^2(Goldstone Phase of
Vacuum Coherence),u,u is ordinary partial derivative.Compare
my equation here to Goldstein's eq. 12.4 on p. 282.PZ:
Consequently,uni?d g_uv has no deep physical meaning or
necessity -- as Feynman argued.PZ: Which all seems to
contradict his statement,...Of course, nothing [in GR] to
prevents us... from singling out thegravitational ?ld as
?themore equal among equals', and declaring that location is
absolute inGR, because it can bede?ed with respect to it.JS:
No, you have simply misread the context of Rovelli's remark.
Readit again more carefully. This is not a valid objection at
all toRovelli's argument IMHO. The way I read him, his
argument issplendidly globally self-consistent.PZ: The way I
read him, he doesn't actually *have* an argument. He simply
*prefers*to quantize space and time intervals rather than
unpack Einstein's uni?d g_uv.PZ: So IMO Rovelli's 
position
is incoherent. If anything, he isarguing for an
*alternative*non-Einsteinian interpretation of GR in which
matter-dependent g_uvis to be treated as a*physical ?ld*,
which means it can in principle be distinguishedfrom the
kinematical inertial?ld with its trivially valid
transformable metric tensor.JS: We are back to Square One.
The Rock has rolled back down onSisyphus. I cannot pinpoint
the precise misconception that is drivingyour position
here.PZ: It is no wonder you are having this dif?ulty, since
it is *Rovelli's* misconceptionthat is causing the
problem.Rovelli doesn't seem to understand that recognition
of the dynamical character ofnon-inertial g_uv fundamentally
distinguishes it from the kinematical g_uv, whichlatter
doesn't depend on the distribution of matter. So the natural
thing to do isto split uni?d g_uv and thus clearly separate
the dynamical and kinematical metric?lds.JS: Show me an
algorithm to do that in any problem.PZ: Yet he doesn't even
consider this.In other words, Rovelli is really a
*Yilmazian*.JS: Not in my book.PZ: Actually, you are right --
because he does not follow the natural developmentof his own
argument to its logical conclusions. If he did, his ?st
alternative (p 109)would fall squarely in the Feynman-Yilmaz
category, with rubber rods and clocksand a ?ckground
inertial metric.JS: It seems to me the objective is to
eliminate any ?ckground non-dynamical
metric.Nondynamical means ACTION WITHOUT REACTION.Note, I
have it formally in my equations but it is not physically
measurable separately.Indeed, the pre-in?ary vacuum is
the perfectly ?rld crystal with no vortex string
topologicaldefects of effective multiple-connectivity in the
space of the vacuum coherence order parameter whichis zero
everywhere prior to the in?ary vacuum phase transition
like a normal metal going super-conducting.PZ: Obviously it
is a ?ld that depends on the distribution ofmatter, if
that's what hemeans. But it is the only ?ld that is de?ed
in terms of aspacetime metric, so itis obviously unique in
that sense.JS: Read Rovelli's paper in the book. Then tell me
what you think.PZ: I read it. Several times. I've been
scratching my head for severaldays now.See above.
===
> I
can't ?d a proof of the (multivariable) chain rule that
makes sense.Assume g is differentiable at x and f is
differentiable at g(x), where differentiable means in the
total derivative sense. Let dg(x) and df(g(x)) denote the
associated linear differential approximants. Then f(g(x+h)) -
f(g(x)) = df(g(x))[g(x+h)) - g(x)] + o(g(x+h)) - g(x)) =
df(g(x))[dg(x)(h) + o(h)] + o(g(x+h)) - g(x)) =
df(g(x))[dg(x)(h)] + df(g(x))[o(h)] + o(g(x+h)) - g(x)).The
?st term in the last sum is what we want to see, namely the
composition of df(g(x)) and dg(x) evaluated at h. So we're
done if we can show both df(g(x))[o(h)] and o(g(x+h)) - g(x))
are o(h), which is not too hard.
===
> I can't ?d a proof of
the (multivariable) chain rule that makes sense.Assume g is
differentiable at x and f is differentiable at g(x), where
>differentiable means in the total derivative sense. Let
dg(x) and >df(g(x)) denote the associated linear differential
approximants. Then f(g(x+h)) - f(g(x)) = df(g(x))[g(x+h)) -
g(x)] + o(g(x+h)) - g(x)) = df(g(x))[dg(x)(h) + o(h)] +
o(g(x+h)) - g(x)) = df(g(x))[dg(x)(h)] + df(g(x))[o(h)] +
o(g(x+h)) - g(x)).The ?st term in the last sum is what we
want to see, namely the >composition of df(g(x)) and dg(x)
evaluated at h. So we're done if we can >show both
df(g(x))[o(h)] and o(g(x+h)) - g(x)) are o(h), which is not
too >hard.Fine, but she said I want to see a real proof that
uses de?itions and little greek letters...I suppose we can
leave it to the reader to change a few lettersto greek to
make things easier to follow. But if she's never seenthis
proof then possibly she's never seen the de?ition:(Def: o(h)
means some function e(h) with the propertythat e(h)/||h|| -> 0
as h -> 0.)Def: f: R^n -> R^m is _differentiable_ at x if
there exists alinear map T: R^n -> R^m such that f(x + h) =
f(x) + Th + o(h);if so then T = df(x).(Then we should also
point out how the notation she maybe looking at follows from
this: If df(x) exists then the matrixcorresponding to that
linear map has the partial derivativesof the components of f
for its entries, and the formula yousee for the chain rule in
some calculus books is justgiving the product of two
matrices.)************************David C. Ullrich
===
> I
can't ?d a proof of the (multivariable) chain rule that makes
sense.>Assume g is differentiable at x and f is
differentiable at g(x), where >differentiable means in the
total derivative sense. Let dg(x) and >df(g(x)) denote the
associated linear differential approximants. Then> f(g(x+h))
- f(g(x)) = df(g(x))[g(x+h)) - g(x)] + o(g(x+h)) - g(x))> =
df(g(x))[dg(x)(h) + o(h)] + o(g(x+h)) - g(x))> =
df(g(x))[dg(x)(h)] + df(g(x))[o(h)]> + o(g(x+h)) -
g(x)).>The ?st term in the last sum is what we want to see,
namely the >composition of df(g(x)) and dg(x) evaluated at h.
So we're done if we can >show both df(g(x))[o(h)] and
o(g(x+h)) - g(x)) are o(h), which is not too >hard.Fine, but
she said I want to see a real proof that uses > de?itions and
little greek letters...> How many students have you found
who've said that? Not many?It reads to me as though this
person is taking a second year course at anAmerican
University. The text books taught from can be pretty dire
when itcomes to formal proof. Indeed one teaches three
?different' (at least) chainrules for special cases, rather
than just _the_ chain rule.I'm sure that the course
instructor would be happyto provide some way for the OP to
satisfy her curiosity, even if it'spointing her towards the
library and a relevant book. But just in case, asyou're the
Analyst round here, which ones would be useful in
demonstratingthat actually you don't want too many greek
letters ?g around? IsRudin any good? Lot's of grad
programs seem to prefer it.> I suppose we can leave it to the
reader to change a few letters> to greek to make things easier
to follow. But if she's never seen> this proof then possibly
she's never seen the de?ition:(Def: o(h) means some function
e(h) with the property> that e(h)/||h|| -> 0 as h -> 0.)Def:
f: R^n -> R^m is _differentiable_ at x if there exists a>
linear map T: R^n -> R^m such that f(x + h) = f(x) + Th +
o(h);if so then T = df(x).(Then we should also point out how
the notation she may> be looking at follows from this: If
df(x) exists then the matrix> corresponding to that linear
map has the partial derivatives> of the components of f for
its entries, and the formula you> see for the chain rule in
some calculus books is just> giving the product of two
matrices.)> ************************David C. Ullrich
===
>
> I can't ?d a proof of the (multivariable) chain rule
that makes sense.>Assume g is differentiable at x and f is
differentiable at g(x), where >differentiable means in the
total derivative sense. Let dg(x) and >df(g(x)) denote the
associated linear differential approximants. Then>
f(g(x+h)) - f(g(x)) = df(g(x))[g(x+h)) - g(x)] + o(g(x+h)) -
g(x))> = df(g(x))[dg(x)(h) + o(h)] + o(g(x+h)) - g(x))>
= df(g(x))[dg(x)(h)] + df(g(x))[o(h)]> + o(g(x+h)) -
g(x)).>The ?st term in the last sum is what we want to
see, namely the >composition of df(g(x)) and dg(x)
evaluated at h. So we're done if we can >show both
df(g(x))[o(h)] and o(g(x+h)) - g(x)) are o(h), which is not
too >hard.> Fine, but she said I want to see a real
proof that uses > de?itions and little greek letters...>
>How many students have you found who've said that? Not
many?It reads to me as though this person is taking a second
year course at an>American University. The text books taught
from can be pretty dire when it>comes to formal proof. Indeed
one teaches three ?different' (at least) chain>rules for
special cases, rather than just _the_ chain rule.I'm sure
that the course instructor would be happy>to provide some way
for the OP to satisfy her curiosity, even if it's>pointing her
towards the library and a relevant book. But just in case,
as>you're the Analyst round here, which ones would be useful
in demonstrating>that actually you don't want too many greek
letters ?g around? Um, all the comments about greek
letters have been tongue in cheek.I don't know what text
would be useful for demonstrating that youdon't want too many
greek letters ?g around, because theidea that you don't
want too many greek letters ?g aroundis news to me.I
suspect that you were being facetious, actually asking
aboutdemonstrating something else. But I can't ?ure out what
thesomething else might be.>Is>Rudin any good? Lot's of grad
programs seem to prefer it.> I suppose we can leave it to
the reader to change a few letters> to greek to make things
easier to follow. But if she's never seen> this proof then
possibly she's never seen the de?ition:> (Def: o(h)
means some function e(h) with the property> that e(h)/||h||
-> 0 as h -> 0.)> Def: f: R^n -> R^m is _differentiable_
at x if there exists a> linear map T: R^n -> R^m such that>
> f(x + h) = f(x) + Th + o(h);> if so then T = df(x).>
> (Then we should also point out how the notation she may>
be looking at follows from this: If df(x) exists then the
matrix> corresponding to that linear map has the partial
derivatives> of the components of f for its entries, and the
formula you> see for the chain rule in some calculus books is
just> giving the product of two matrices.)>
************************> David C.
Ullrich************************David C. Ullrich
===
Need help
on math, visit www.helptosolve.com Just try.Want to help
others with math, visit www.helptosolve.com and join the
team.
===
>Dean said that Northeasterners don't talk about
religion. >That's not true, because I know of
Northeasterners who spend almost>all day talking about The
Virgin Mary, Hearing Confession, who's Rabbi>is where, what
Temple this person goes to, I have another Bar
Mitzvah,>Episcopal Parishoners this, American Baptists
convention that.>What Northeast is Dean from ??> What planet
are you from?We have people from every planet on Earth in
California. --- Formergovernor Gray Davis--
http://hertzlinger.blogspot.com
===
> Many of the problems
with Bush's agenda is the dishonest> reinterpretation of
reality that is done so often by the> Democrats.Citations,
please.
===
> Dean said that Northeasterners don't talk about
religion. That's not true, because I know of Northeasterners
who spend almost> all day talking about The Virgin Mary,
Hearing Confession, who's Rabbi> is where, what Temple this
person goes to, I have another Bar Mitzvah,> Episcopal
Parishoners this, American Baptists convention that.What
Northeast is Dean from ??What does this have to do with
mathematics?-- http://hertzlinger.blogspot.comX-Cise:
tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate:
admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose:
george_cox@btinternet.comX-Punge: Micro$oft
===
>Dean said
that Northeasterners don't talk about religion. K3wl? How is
that on topic for sci.math, sci.skeptic, alt.atheism
oralt.christnet? Of the 5 groups that you posted to,
onlyalt.politics.democrats is relevant; clearly your intent
was to trollfor a crss-posted ?ar.-- Shmuel (Seymour J.)
Metz, SysProg and JOATnot reply to
spamtrap@library.lspace.org
===
> Dean said that
Northeasterners don't talk about religion. Maybe you should
furnish an actual quote before you build a castle in the >
air upon what you think he might have said. ?'My father used
to tell us how much strength he got from religion, but > we
didn't have Bible readings. There are traditions where people
do > that. We didn't,'' he said. 
?'People in the Northeast
don't talk about > their religion. It's a very 
personal
private matter, and that's the > tradition I was brought up
in.'' Howard Dean, to Boston Globe last > week. Quoted 
in
countless other papers.You gotta interpret that in context.
It means northeastern politiciansdon't talk to the papers
about their religion (they leave that to preachers). It was
de?itely more polite than what I would have said, which
would be, I didn't come here today to talk about religion, so
what's your next question?
===
> Dean said that
Northeasterners don't talk about religion. > Maybe you
should furnish an actual quote before you build a castle in
the > air upon what you think he might have said. > ?'My
father used to tell us how much strength he got from religion,
but > we didn't have Bible readings. There are traditions
where people do > that. We didn't,'' he 
said. ?'People in
the Northeast don't talk about > their religion. It's 
a very
personal private matter, and that's the > tradition I was
brought up in.'' Howard Dean, to Boston Globe last > 
week.
Quoted in countless other papers.You gotta interpret that in
context. It means northeastern politicians>don't talk to the
papers about their religion (they leave that to >preachers).
It was de?itely more polite than what I would have said,
>which would be, I didn't come here today to talk about
religion, so >what's your next question?By the same logic the
previous poster used, Bush has also told lies.President Bush's
compassionate agenda resonates with the people ofboth New York
and California, -Tracey Schmitt,spokeswoman for the
Bush-Cheney ?04campaignThat's not true, because I know of
Californians who are not impressedwith Bush's compassionate
agenda. -----Yang a.a. #28a.a. pastor #-273.15, the most
frigid church of Celcius nee Kelvin EAC Econometric Forecast
and Socerey DivisionProudly plonked by Lani Girl and
CrazyalecThe Bush ?balanced' budget: -525 billion and
worseningThe Bush ?economic' policy: -3 million jobs and
countingThe Bush Iraq lie: -472 GIs, one friend's 
co-worker's
son and mountingHaving Bush  up my country: Worthless
===
>
Dean said that Northeasterners don't talk about religion.We
don't, certainly not when compared to other parts ofthe
country.Most people I know consider it an impolite subject,
atleast in public & amongst strangers.
===
 > Dean said that
Northeasterners don't talk about religion.> Maybe you
should furnish an actual quote before you build a castle
inthe> air upon what you think he might have said.> ?'My
father used to tell us how much strength he got from
religion, but> we didn't have Bible readings. There are
traditions where people do> that. We 
didn't,'' he said.
?'People in the Northeast don't talk about> their 
religion.
It's a very personal private matter, and that's the>
tradition I was brought up in.'' Howard Dean, to 
Boston Globe
last> week. Quoted in countless other papers.I heard my father
once say that people in the South arevery conservative
religiously. Well I live in Atlanta nowand am friends with
some folk at the Unitarian Church.My father. What a dirty,
rotten liar.
===
|Did you see my reply to the OP?Yes, although
the contents were not still clear in mind when I lastAmanda's
acquaintance objected to her proof based on its being aproof
by contradiction. I guess I think the distinction between
aproof by proving the contrapositive and a proof by
contradiction isfuzzy enough not to make this strictly
incorrect. For the questionof whether the proof is valid, I
think you'd agree it's also not adistinction that 
could be
relevant. People consider the proof okaynot because it's
crossed this fuzzy line; people consider the proofokay
because we're considering a context where proof by
contradictionand the whole complex of related forms of
reasoning are all accepted.You had two main reasons why proof
by contradiction was consideredworse:|(i) it's easy to give a
wrong proof by contradiction, where |the contradiction arises
from some error ||(ii) if you give a direct proof that A
implies B, by assuming|A and then deducing B, the steps in
the proof can give|some insight into what A really entails -
you show A implies|B by showing A implies C and C implies B,
and along the|way you've shown two facts that might be
interesting and|useful elsewhere, that A implies C and that C
implies B.|You don't get this sort of bonus from a proof by
contradiction,|since in the course of the proof you're
assuming things|which it turns out never actually hold.I
think the practice of constructive proof may shed some light
onthis. The conclusion of a constructive proof tells you more
aboutthe nature of the proof than the conclusion of a
classical proof,generally. In particular, when the conclusion
is negative, the proofwas a proof by contradiction
(essentially) and when the conclusionis positive it wasn't.I
can't tell whether the circumstances reducing the risk of (i)
andenhancing the prospects of (ii) are different from the
circumstanceswhere the theorem (as it would be stated in
constructive terms) ismore solidly stated.In constructive
mathematics one has the general advice of Bishopto cut down
on negations. If we seek to tighten up our theorems bygetting
rid of ?negations, then we also as a side-effectrefrain
from proofs by contradiction.Lots of negations and
implications can be upgraded by supplying themwith more solid
content. Bishop has a paper which uses Goedel'sDialectica
paper for this. Goedel has a method of rewritingstatements
which puts them into a form (there exists w) (for all t)
P(w,t)where w ranges over a set of possible witnesses to the
truth ofthe statement, and t ranges over a set of possible
cases. If thisstatement is false, one would like to do more
than just negate it;one would like to say something about the
function f:w->t for whichP(w, f(w)) fails for all w. A proof
of the negative statementimplicitly provides such an f, as
witness. Even when a statementis negative, then, there can be
implicit solid content in theconstruction demonstrating the
fact, in terms of information onand around this f.I get the
impression that in cases where a theorem is negative,but has
relatively tangible constructive content, one is not
soespecially prone to err in the way described in (i),
because oneis regarding the construction in the proof (of f,
say), whichactually *does* exist, and trying to see whether
*it* is asadvertised. In the proof that there are in?itely
many primes,to pick a simple example, the engine is the
method for gettinga new prime out of a ?ite set of old
primes. An argument whichfocusses on this kind of concrete
construction doesn't seem tosuffer from the kind of problem
that some attempts at proof bycontradiction do.I'm going on
impressions, though, and it seems like it wouldtake somewhat
tricky casework to get a more solid sense for whatkinds of
proof do and do not tend to suffer from these problems.Keith
Ramsay
===
|Did you see my reply to the OP?Yes, although the
contents were not still clear in mind when I lastAmanda's
acquaintance objected to her proof based on its being a>proof
by contradiction. Although I don't think that even he was
saying the proof wastherefore _wrong_, just that it was not
the best possible proof.(Cf for example the title of the
thread; it's about ?ding abeautiful proof, not a _valid_
proof...)>I guess I think the distinction between a>proof by
proving the contrapositive and a proof by contradiction
is>fuzzy enough not to make this strictly incorrect. I'm
honestly not sure what you mean by that sentence,
butregarding the minor premise, I don't see what's 
fuzzy
aboutthe distinction. To prove A implies B:(a) Assume ~B.
Prove ~Aor(b) Assume A and ~B. Prove P and ~P for some P.Some
proofs of A implies B have the form (a); somehave the form
(b). What's fuzzy here?Oh. You're claiming that the
distinction is fuzzy enoughthat those of us who are saying he
was wrong instating the proof was a proof by contradiction are
noton solid ground? I disagree. Um, lemme put that alittle
more strongly: that's not so. The proof Amandagave is of the
form (a), not of the form (b).I mean that's just a
mathematical _fact_. You can sayyou don't see why it matters,
?e. But you say aboveyou think the distinction between (a)
and (b) isfuzzy - I don't see any fuzziness, and it's 
simply
afact that (a) is an outline of the proof she gave,
forappropriate A and B, while (b) simply isn't.>For the
question>of whether the proof is valid, I think you'd agree
it's also not a>distinction that could be relevant. People
consider the proof okay>not because it's crossed this fuzzy
line; people consider the proof>okay because we're
considering a context where proof by contradiction>and the
whole complex of related forms of reasoning are all
accepted.Yes of _course_. Of the people in this thread who do
see the difference between (a) and (b) above, some of whom
have beensaying that (a) is preferable, _none_ of them have
stated oreven hinted that (b) is invalid.>You had two main
reasons why proof by contradiction was considered>worse:|(i)
it's easy to give a wrong proof by contradiction, where >|the
contradiction arises from some error >|>|(ii) if you give a
direct proof that A implies B, by assuming>|A and then
deducing B, the steps in the proof can give>|some insight
into what A really entails - you show A implies>|B by showing
A implies C and C implies B, and along the>|way you've shown
two facts that might be interesting and>|useful elsewhere,
that A implies C and that C implies B.>|You don't get this
sort of bonus from a proof by contradiction,>|since in the
course of the proof you're assuming things>|which it turns
out never actually hold.I think the practice of constructive
proof may shed some light on>this. The conclusion of a
constructive proof tells you more about>the nature of the
proof than the conclusion of a classical proof,>generally. In
particular, when the conclusion is negative, the proof>was a
proof by contradiction (essentially) and when the
conclusion>is positive it wasn't.I can't tell whether 
the
circumstances reducing the risk of (i) and>enhancing the
prospects of (ii) are different from the circumstances>where
the theorem (as it would be stated in constructive terms)
is>more solidly stated.In constructive mathematics one has
the general advice of Bishop>to cut down on negations. If we
seek to tighten up our theorems by>getting rid of ?egat
ions, then we also as a side-effect>refrain from proofs
by contradiction.Lots of negations and implications can be
upgraded by supplying them>with more solid content. Bishop
has a paper which uses Goedel's>Dialectica paper for this.
Goedel has a method of rewriting>statements which puts them
into a form (there exists w) (for all t) P(w,t)where w ranges
over a set of possible witnesses to the truth of>the
statement, and t ranges over a set of possible cases. If
this>statement is false, one would like to do more than just
negate it;>one would like to say something about the function
f:w->t for which>P(w, f(w)) fails for all w. A proof of the
negative statement>implicitly provides such an f, as witness.
Even when a statement>is negative, then, there can be implicit
solid content in the>construction demonstrating the fact, in
terms of information on>and around this f.I get the
impression that in cases where a theorem is negative,>but has
relatively tangible constructive content, one is not
so>especially prone to err in the way described in (i),
because one>is regarding the construction in the proof (of f,
say), which>actually *does* exist, and trying to see whether
*it* is as>advertised. In the proof that there are in?itely
many primes,>to pick a simple example, the engine is the
method for getting>a new prime out of a ?ite set of old
primes. An argument which>focusses on this kind of concrete
construction doesn't seem to>suffer from the kind of problem
that some attempts at proof by>contradiction do.I'm going on
impressions, though, and it seems like it would>take somewhat
tricky casework to get a more solid sense for what>kinds of
proof do and do not tend to suffer from these problems.The
context here, or at least the context I had in mind,
wasproofs by students in classes where half the point to the
classis learning to read and write proofs (beginning
algebra,analysis, topology classes are often in this
category).In that context I can tell you from experience that
(i)is a real danger. Of course it's much less of a problemwhen
we're talking about proofs in the real world writtenby
grownups - one has an idea what _sort_ of contradictionto
expect, so when one gets a totally irrelevant
contradictionone looks for the error instead of saying
qed.>Keith Ramsay************************David C.
Ullrich
===
> For the greater cogency and obviousness in
your paper THERE> SHOULD BE a TABLE with componentries of
system, which one are> sorted out on their in? to
precision of system GPS as a whole.Such a table was posted by
Sam Wormley on the 22nd, seeThen any layman can see that we
can neglect neglible small> relativistic corrections as
contrasted to by other factors> de?ing and restricting
limiting accuracies GPS as a whole.> VI. Summary> Excluding
the deliberate degradation of SA, the dominant error source>
for satellite ranging with single frequency receivers is
usually the> ionosphere. It is on the order of four meters,
depending on the> quality of the single-frequency model. For
dual-frequency (P/Y-code)> receivers (which eliminate SA) the
Standard Error Model of Table I> has one principal change (in
addition to the elimination of the SA> error). The
ionospheric error is reduced from four meters to about> one
meter.> The GR correction is about 44 microseconds or 13km
per day> and would be cumulative. You can hardly call that
negligible.1. What periodicity of corrections of parameters
in GPS? 2. What parameters are adjusted in GPS? 3. What
medial - statistical values have parameters adjusted in GPS
in each session of corrections?--Aleksandr
===
> For the
greater cogency and obviousness in your paper THERE> SHOULD
BE a TABLE with componentries of system, which one are>
sorted out on their in? to precision of system GPS as a
whole.> Such a table was posted by Sam Wormley on the 22nd,
see> Then any layman can see that we can neglect neglible
small> relativistic corrections as contrasted to by other
factors> de?ing and restricting limiting accuracies GPS as
a whole.> VI. Summary> Excluding the deliberate
degradation of SA, the dominant errorsource> for satellite
ranging with single frequency receivers is usually the>
ionosphere. It is on the order of four meters, depending on
the> quality of the single-frequency model. For
dual-frequency (P/Y-code)> receivers (which eliminate SA)
the Standard Error Model of Table I> has one principal
change (in addition to the elimination of the SA> error).
The ionospheric error is reduced from four meters to about>
one meter.> The GR correction is about 44 microseconds or
13km per day> and would be cumulative. You can hardly call
that negligible. 1. What periodicity of corrections of
parameters in GPS?The corrections are of the order of
nanoseconds per day socon?m GR to roughly one part in 10,000
by that simplisticcomparison. There is much more information
available in webpages if you want to look for more details.>
2. What parameters are adjusted in GPS?> 3. What medial -
statistical values have parameters adjusted> in GPS in each
session of corrections?Happy New YearGeorge
===
To base
10:STEP 1: For the characteristic: carry out repeated
division of the number by10 until the result falls below 10
and count the number of times.STEP 2: For the mantissa: raise
the result to the power of 10 usingmultiplication only, then
divide by 10 again as in Step 1. Continue to Step2 for
further digits.This process generates the logarithm digit by
digit, to the same accuracy asthe other arithmetical
functions. Also works ef?iently in binary.This has almost
certainly been tried before. I would be interested in
any
===
equation.The equation isAx=0A is a very large sparse
matrix which have 1473-by-1473.Sum(x_i) is 1.0All x are not
zero.Please, would you like to let me know how to solve the
simultaneousequation by fortran.Additionally, My major is not
mathmatics. Thus, it is dif?ult tosolve the equation in a
method of pseudoinverse, SVD, etc.. Please,let me know that,
easily and in detail.
===
> equation. The equation is Ax=0 A
is a very large sparse matrix which have 1473-by-1473.>
Sum(x_i) is 1.0> All x are not zero. Please, would you like
to let me know how to solve the simultaneous> equation by
fortran. Additionally, My major is not mathmatics. Thus, it
is dif?ult to> solve the equation in a method of
pseudoinverse, SVD, etc.. Please,> let me know that, easily
and in detail.>Well, I haven't dealt with matrices that
large, but maybe you could tryCramer's rule. As for the code,
you are on your own, I don't know Fortran.You could
try:http://www.library.cornell.edu/nr/bookfpdf.htmlLurch
===
>
The equation is> Ax=0> A is a very large sparse matrix which
have 1473-by-1473.>Well, I haven't dealt with matrices that
large, but maybe you could try>Cramer's rule. You know, I
think all of us have a tendency to think we can answermore
questions well than we really can. I don't know a whole
heckuva lotabout numerical analysis, but this strikes me as a
really, really badanswer. For one thing, Cramer's rule gives
an explicit solution toAx=b when A is invertible; the OP's
matrix A is surely notinvertible, lest the only solution be x
= 0. For another thing,Cramer's rule expresses the answer in
terms of (here) 1474 determinantsof 1473x1473 matrices. How
do you propose that the determinants beevaluated? A naive
algorithm sums 1473! products and introduces thepossibility
of tremendous quantities of round-off error and
massivecancellations. Even using row operations, while much
faster, is stilltime consuming and prone to error if the
entries are not, say, small integers. As useful as Cramer's
rule can be for theoretical work,I imagine it's essentially
_never_ used for computing numerical solutionsto linear
equations with more than a handful of variables.So what
response might be better for the OP?It's probably true that
any good matrix package can handle yourmatrix (a couple
thousand rows is not considered huge these days)but since you
say your matrix is sparse, you would probably bene?from using
routines speci?ally written for this common special case.Here
is a link which was posted the other day in
sci.math.num-analysis(which is a much more appropriate group
for technical questionsof this type):
http://vlsicad.cs.ucla.edu/sparse.htmlOf course the equation
Ax=0 has no solution with sum(x_i)=1 if A isnonsingular, so
it would be good to think about how you know thatyour problem
has a solution. More generally, it is likely to helpthose who
are helping you if you can indicate where your matrix
comesfrom, since that may signal the use of special routines
e.g. forsolving differential equations numerically.dave
===
>
> The equation is> Ax=0> A is a very large sparse
matrix which have 1473-by-1473. >Well, I haven't dealt with
matrices that large, but maybe you could try>Cramer's rule.
You know, I think all of us have a tendency to think we can
answer> more questions well than we really can. I don't know
a whole heckuva lot> about numerical analysis, but this
strikes me as a really, really bad> answer. For one thing,
Cramer's rule gives an explicit solution to> Ax=b when A is
invertible; the OP's matrix A is surely not> invertible, lest
the only solution be x = 0. For another thing,> Cramer's rule
expresses the answer in terms of (here) 1474 determinants> of
1473x1473 matrices. How do you propose that the determinants
be> evaluated? A naive algorithm sums 1473! products and
introduces the> possibility of tremendous quantities of
round-off error and massive> cancellations. Even using row
operations, while much faster, is still> time consuming and
prone to error if the entries are not, say,> small integers.
As useful as Cramer's rule can be for theoretical work,> I
imagine it's essentially _never_ used for computing numerical
solutions> to linear equations with more than a handful of
variables. So what response might be better for the OP? It's
probably true that any good matrix package can handle your>
matrix (a couple thousand rows is not considered huge these
days)> but since you say your matrix is sparse, you would
probably bene?> from using routines speci?ally written for
this common special case.> Here is a link which was posted
the other day in sci.math.num-analysis> (which is a much more
appropriate group for technical questions> of this type):>
http://vlsicad.cs.ucla.edu/sparse.html Of course the equation
Ax=0 has no solution with sum(x_i)=1 if A is> nonsingular, so
it would be good to think about how you know that> your
problem has a solution. More generally, it is likely to help>
those who are helping you if you can indicate where your
matrix comes> from, since that may signal the use of special
routines e.g. for> solving differential equations
numerically. daveHey, it's an open forum! I consider this NG
like an online math club. Ioffer suggestions, and they aren't
always useful, or correct. None of thisstuff is being
published, so relax. He isn't paying me for my help; so, heis
free to take my advice, or not. It is up to him. I also
provided a linkto an online book of Numerical analysis in
Fortran, which should answer hisquestion.Lurch
===
> The
equation is> Ax=0> A is a very large sparse matrix
which have 1473-by-1473.>Well, I haven't dealt with matrices
that large, but maybe you could try>Cramer's rule. You know,
I think all of us have a tendency to think we can answer>more
questions well than we really can. [...]I have an explanation
for that... hmm, no I don't.************************David C.
Ullrich
===
I am looking for a formula for the calculation of
the midpoint for thesmallest, an area A circumscribing
circle. Is there like for trianglesa connection between the
center of gravity and the center of thatcircle? Are any
formulas for such a circle known?Carolin
HauÛner
===
hiwikipedia needs our help to carry
on:http://www.wikimedia.org/letter.htmlthanks to
allbyemax
===
What's happened to you ?I get up early every
morning to read your diatribes and I haven't seen anything
for a few days now.Does this mean I have to learn maths if I
want to be a part of this newsgroup ?Does this mean I have to
rely on comics for my cackles each morning ?I hope the FBI
didn't turn on you and that the Evil Mathematical
Establishment (tm) haven't imprisioned you in an in?ite
series !Please hurry back, my humour is suffering and my
education lacking - at least I was being *forced* to learn
maths while you posted for no other reason than trying to
understand the people who replied to you!Ivan.So this poster
who's opinion, by his own admissions in terms ofmathematics
is worthless, still feels that his opinion is of
value,clearly because he's at least learned the true nature
of the mathcommunity. JSH writing about me in the Focus on
point of dispute, more math thread ... I've been villi?d by
James, does this make me famous?
===
the application of
nonlinear systems (particularly catastrophetheory) to human
behavior. Through the years I have struggled withbalancing
the criticisms of catastrophe methodology and the
heuristicmerits of the theory. I've read most of the
catastrophe literature andtheory and singularities.Although
the proponents of catastrophe theory are also
somewhatconvincing, I question its applicability and
implementation inpsychological research. I'm at a point where
I need to make adecision:Do I drop this line of research or do
I continue working toward thedevelopment of catastrophe
applications (e.g., write grants to developsoftware for
analyzing catastrophes/bifurcations in
psychologicaldata)?Kate
===
the application of nonlinear
systems (particularly catastrophe> theory) to human behavior.
Through the years I have struggled with> balancing the
criticisms of catastrophe methodology and the heuristic>
merits of the theory. I've read most of the catastrophe
literature and> theory and singularities.Although the
proponents of catastrophe theory are also somewhat>
convincing, I question its applicability and implementation
in> psychological research. I'm at a point where I need to
make a> decision:Do I drop this line of research or do I
continue working toward the> development of catastrophe
applications (e.g., write grants to develop> software for
analyzing catastrophes/bifurcations in psychological>
data)?KateIt would seem that this is more a question for your
advisor/committee/faculty to help you answer. IMO, as a grad
student it is generallyadvisable to do something that gives
you a high probability ofproducing a respectable dissertation
ASAP. A speculative ?hingexpedition (i.e. one where you're
not sure there are any ?h in thelake as opposed to the
standard risk of just not catching any of ?hthere are) is
not what you should undertake early in your career.However,
if you still want to pursue it, ask around about
fundingprospects, that is is anyone supporing this type of
work? Maybethere is all kinds of homeland securtiy money
being thrown atthis area in an attempt to ?ure out
terrorists or something (don'tlaugh, they're spending
billions on a missle defense system thathasn't worked yet).
This is where faculty should be able, throughtheir
connections, to help you with information on research
andfunding trends. But unless your advisor knows about this
area andis supportive of the idea, you're asking for
problems, IMO.Good luck,Russell
===
> the application of
nonlinear systems (particularly catastrophe> theory) to human
behavior. Through the years I have struggled with> balancing
the criticisms of catastrophe methodology and the heuristic>
merits of the theory. I've read most of the catastrophe
literature and> theory and singularities. Although the
proponents of catastrophe theory are also somewhat>
convincing, I question its applicability and implementation
in> psychological research. I'm at a point where I need to
make a> decision: Do I drop this line of research or do I
continue working toward the> development of catastrophe
applications (e.g., write grants to develop> software for
analyzing catastrophes/bifurcations in psychological> data)?
KateWhy study pseudo-science anyway?
===
} Why study
pseudo-science anyway?All the better to refute asinine
one-liners, my dear.
===
> } Why study pseudo-science anyway?
All the better to refute asinine one-liners, my dear.>Are you
claiming it isn't a pseudo-science?Lurch
===
the application of
nonlinear systems (particularly catastrophe> theory) to human
behavior. Through the years I have struggled with> balancing
the criticisms of catastrophe methodology and the heuristic>
merits of the theory. I've read most of the catastrophe
literature and> theory and singularities.> Although the
proponents of catastrophe theory are also somewhat>
convincing, I question its applicability and implementation
in> psychological research. I'm at a point where I need to
make a> decision:> Do I drop this line of research or do I
continue working toward the> development of catastrophe
applications (e.g., write grants to develop> software for
analyzing catastrophes/bifurcations in psychological> data)?>
> KateWhy study pseudo-science anyway?Which one?
;-)Russell
===
> Does the natural density of all positive
integers that arethe sum of two odd> primes = 1/2? How about
the sum of two squares? The sum oftwo squarefree>
integers?**************************************************If
the Goldbach conjecture is true, then the answer to your?st
question is yes.If a number is the sum of two squares then it
can't have aprime factor that is 3 modulo 4 and whose highest
power isodd. Since the series 1 + 1/3 + 1/7 + 1/11 + 1/19 + .
. .diverges, this would imply that the natural density of
thenumbers that are the sum of two squares is 0. ( [
(1+1/3)(1+1/7)(1+1/11) . . . ] ^ (-1) =
0.)_________________________________________________________
Eric J. Wingler (wingler@math.ysu.edu)Dept. of Mathematics
and StatisticsYoungstown State UniversityOne University
PlazaYoungstown, OH 44555-0001330-941-1817
===
> Does the
natural density of all positive integers that are>the sum of
two odd> primes = 1/2? How about the sum of two squares? The
sum of>two squarefree>
integers?**************************************************>
If the Goldbach conjecture is true, then the answer to
your>?st question is yes.If a number is the sum of two
squares then it can't have a>prime factor that is 3 modulo 4
and whose highest power is>odd. Since the series 1 + 1/3 +
1/7 + 1/11 + 1/19 + . . .>diverges, this would imply that the
natural density of the>numbers that are the sum of two squares
is 0.> ( [ (1+1/3)(1+1/7)(1+1/11) . . . ] ^ (-1) = 0.)I give
up. Supposing that everything you say is true, whichI imagine
it is, how does it follow that the sums of two squareshave
density 0?(My guess is that you deduce somehow that the sum
of thereciprocals of the sums of two squares is ?ite. If so
that'sinteresting because the sum of 1/(j^2 + k^2) is
in?ite...)__________________________________________________
_______Eric J. Wingler (wingler@math.ysu.edu)>Dept. of
Mathematics and Statistics>Youngstown State University>One
University Plaza>Youngstown, OH
44555-0001>330-941-1817>************************David C.
Ullrich
===
>If a number is the sum of two squares then it
can't have a>prime factor that is 3 modulo 4 and whose
highest power is>odd. Since the series 1 + 1/3 + 1/7 + 1/11
+ 1/19 + . . .>diverges, this would imply that the natural
density of the>numbers that are the sum of two squares is
0.> ( [ (1+1/3)(1+1/7)(1+1/11) . . . ] ^ (-1) = 0.)>I give
up. Supposing that everything you say is true, which>I
imagine it is, how does it follow that the sums of two
squares>have density 0?Let S(p) be the set of positive
integers that are either not divisible by por are divisible
by p^2. The natural density of S is 1 - 1/p + 1/p^2.Any
positive integer that is the sum of two squares must be in
S(p) forall primes p = 3 mod 4. If these primes are p_1, p_2,
p_3, ..., thenatural density of intersection_{j=1}^n S(p_j) is
product_{j=1}^n (1 - 1/p_j + 1/p_j^2) = product_{j=1}^n
exp(-1/p_j) (1+O(1/p_j^2)) <= C exp(-sum_{j=1}^n 1/p_j))which
goes to 0 as n -> in?ity because the series sum_{j=1}^infty
1/p_j diverges.Robert Israel israel@math.ubc.caDepartment of
Mathematics http://www.math.ubc.ca/~israel University of
British Columbia Vancouver, BC, Canada V6T 1Z2
===
>If a
number is the sum of two squares then it can't have a>prime
factor that is 3 modulo 4 and whose highest power is>odd.
Since the series 1 + 1/3 + 1/7 + 1/11 + 1/19 + . .
.>diverges, this would imply that the natural density of
the>numbers that are the sum of two squares is 0.> ( [
(1+1/3)(1+1/7)(1+1/11) . . . ] ^ (-1) = 0.)>I give up.
Supposing that everything you say is true, which>I imagine
it is, how does it follow that the sums of two squares>have
density 0?Let S(p) be the set of positive integers that are
either not divisible by p>or are divisible by p^2. The
natural density of S is 1 - 1/p + 1/p^2.Ah. That's the bit
that hadn't clicked - couldn't see where the 
minussigns in
the expansion of 1/(1+1/p) were going to come in, but thereit
is. Duh.>Any positive integer that is the sum of two squares
must be in S(p) for>all primes p = 3 mod 4. If these primes
are p_1, p_2, p_3, ..., the>natural density of
intersection_{j=1}^n S(p_j) is >product_{j=1}^n (1 - 1/p_j +
1/p_j^2) > = product_{j=1}^n exp(-1/p_j) (1+O(1/p_j^2))> <= C
exp(-sum_{j=1}^n 1/p_j))>which goes to 0 as n -> in?ity
because the series sum_{j=1}^infty 1/p_j >diverges.Robert
Israel israel@math.ubc.ca>Department of Mathematics
http://www.math.ubc.ca/~israel >University of British
Columbia >Vancouver, BC, Canada V6T
1Z2************************David C. Ullrich
===
THe cost of a
long-distance telphone call is determined by a?e for teh
?st 5 minutes and a ?ed amount for each additional minute.
If a 15-minute telephone call costs $3.25 and a 23-minute
call costs $5.17, ?d the cost of a 30-minute call. I was
thinking of setting it up as a system of equations but im not
really sure, can someone explain how i would set this problem
up, cause alwyas get stumped on these type of questions.
TIA---= 19 East/West-Coast Specialized Servers - Total
Privacy via Encryption =---
===
> THe cost of a long-distance
telphone call is determined by a?e for> teh ?st 5
minutes and a ?ed amount for each additional minute. If
a15-> minute telephone call costs $3.25 and a 23-minute call
costs $5.17, ?d> the cost of a 30-minute call.> I was
thinking of setting it up as a system of equations but im
not> really sure, can someone explain how i would set this
problem up, cause> alwyas get stumped on these type of
questions. TIANews==----> http://www.newsfeed.com The #1
Newsgroup Service in the World! >100,000> ---= 19
East/West-Coast Specialized Servers - Total Privacy via
Encryption=---Let x = the ?eand y = the ?ed amount
after the ?st ?e,then5x + 18y = 5.175x + 10y = 3.25You get
x = .17 and y = .24.Lurch
===
> THe cost of a long-distance
telphone call is determined by a?e for> teh ?st 5
minutes and a ?ed amount for each additional minute. If
a>15-> minute telephone call costs $3.25 and a 23-minute
call costs $5.17, ?d> the cost of a 30-minute call.>Let x
= the ?e>and y = the ?ed amount after the ?st
?e>,then5x + 18y = 5.17>5x + 10y = 3.25You get x = .17 and y
= .24.Lurch>No, a ?e is ?ed and pays for anything up to
including 5minutes, not a per minute fee. You should have x
in the equationsinstead of 5x. The ?st ?e minutes cost .85
(the ?e) and theadditional minutes are .24 each. A 2
minute call would also cost .85because you don't apply the
?e on a per minute basis.--Lynn--Lynn
===
 > THe cost of
a long-distance telphone call is determined by a?efor>
teh ?st 5 minutes and a ?ed amount for each additional
minute. If a>15-> minute telephone call costs $3.25 and a
23-minute call costs $5.17,?d> the cost of a 30-minute
call.>Let x = the ?e>and y = the ?ed amount after
the ?st ?e>,then>5x + 18y = 5.17>5x + 10y = 3.25>You
get x = .17 and y = .24.>LurchNo, a ?e is ?ed and
pays for anything up to including 5> minutes, not a per
minute fee. You should have x in the equations> instead of
5x. The ?st ?e minutes cost .85 (the ?e) and the>
additional minutes are .24 each. A 2 minute call would also
cost .85> because you don't apply the ?e on a per minute
basis. --Lynn --Lynn>Let c=cost m=cost per minute over 5 and
x=call lengthc=.85+m(x-5)c=.85+m(x-5)c=.85+mx-5m, x>5As Lynn
said, anything below 5 remains .85A 15 minute call costs
$3.25 so x=15 and c=3.253.25=.85+15m-5m2.4=10mm=$.24So Now we
know m and our formula becomes c=.24x-.35So when x=30, c=6.85.
Therefore a 30 minute call costs $6.85David Moran
===
Can the
equation for c be transformed into an elliptic curve?> The
Barcelona conjecture:> Let c=(x+y+z)^p/(pxyz2^p)> for
integer c,x,y,z and p prime greater than or equal to 5, the>
> Barcelona conjecture is that no solutions exist with
gcd(c,xyz)=1 (no> c exist that shares no factor with x or y
or z).> I haven't seen this conjecture before, but compare the
Beal conjecture:> www.math.unt.edu/~mauldin/beal.htmlYes, I
was aware of the Beal conjecture and brie?empted to
prove> it also :)The reason you hadn't seen the Barcelona
conjecture is that it is> virtually unknown outside of this
newsgroup since I've only posted it> here and in the research
math group. I came up with it while> attempting to prove FLT
using elementary techniques ( ok, once we are> done laughing
the question remains - who hasn't? I mean even JSH keeps> on
trying.)In some respects it should be easier to prove FLT
using the Barcelona> conjecture as the latter places less
restrictions on the value of c -> maybe even Fermat was
working on this approach as it only requires> elementary
methods, though I really doubt it as it isn't documented in>
Ribenboim's book on FLT.If anyone can lend me a hand 
I'm
trying to get a grip on how FLT was> tied to elliptic curves,
my goal being to see how feasible it is to> apply the same
methods to the Barcelona conjecture - for the moment it> is
way too dif?ult for me.> 
===
BTW; did anyone here bother to
check out David Sereda, and of hishttp://www.ufonasa.com/
EVIDENCE the case for NASA UFOsIn some recent additions to my
MAZDA like Internal Rocket RotaryCombustion Engine (IRRCE sfc
= 15+KW/kg), there seems we also haveourselves a wee bit of
lunar He3 to burn
off.http://guthvenus.tripod.com/gv-h2o2-irrce.htmThe taking
of lunar He3 is ?st come ?st served, that could bethe likes
of China or Russia because, we're obviously not
smartenough.Venus still offers life; via moon He3 could help
turn the trickI've got a few more words of wisdom to offer on
behalf of the ARTEMISPROJECT (lunar He3)
http://guthvenus.tripod.com/gv-lse-he3.htmThe rest of this
report isn't entirely related to energy so much as itrelates
to truth or consequences. Such as for this next topic of
therebeing other life NOT as we know it on Venus, that's
obviously opposingthe sorts of anti-humanity folks that
couldn't care less if our entireworld was destroyed by their
resident warlord. Perhaps these folks canget their next level
of future funding from the same source as Bush,Salem Laden.As
for making policy look like happenstance, and/or vice versa,
is keyto snookering folks.
http://guthvenus.tripod.com/moon-04.htmThough as for we
humans need not, and perhaps we should not ventureourselves
much beyond Venus L2 (VL2). Wouldn't want to contaminate
aperfectly good planet with our inferior DNA nor lack of
morals,especially of this group that's bashing honest
research just out ofspite. Besides, their stealth
donkey-carts could be far more lethalthan what our WMD
donkey-carts can manage.As far as our human physiology being
adaptable to pressure. Under suchpressure things are not
nearly as hot as we've been told, and you wontneed but a
fraction of a percent of O2. Of course, that degree
ofadaptation might have to be accommodated at a modus rate of
a few barsper day.http://guthvenus.tripod.com/venus-air.htmI
have a few other recent/ongoing comments on H2O2/C12H26 and
of
He3:http://guthvenus.tripod.com/gv-irrce.htmhttp://
guthvenus.tripod.com/gv-hybrid-irc.htmhttp://
guthvenus.tripod.com/gv-cm-ccm-01.htmhttp://
guthvenus.tripod.com/gv-lm-1.htmhttp://guthvenus.tripod.com/
radio-maybe.htm
===
Portfolio of PAF as of 28DEC03BCE 400
21.90 $8,760.00BLS 50 27.92 $1,396.00BMY 100 27.80 $2,780.00
Q 50,000 3.92 $196,000.00SBC 11,600 25.68
$297,888.00realestate land 3APR03 of 3 lots
$19,000science-art of pictures,porcelain etc starting JAN03
for $12,160.realestate land 30JUL03 another lot $11,500.and
Wyeth at a small pro? and with the proceeds bought 200
moreshares of BCE.Sold DT because it is hard to play DT as a
Crossover switchingcampaign. Sold Verizon because it is
overpriced compared to SBC. Andsold Wyeth because I was never
really aware of the history of phenfen (excuse the spelling)
until I saw a PBS program on that drug andthe history of the
company of Wyeth in relation to that drug. Afterseeing the
history of phen fen, I could not help but think that Wyethis
a company that fosters a culture of money grubbing ever more
thanit fosters a culture that they would seek health
medicines to makethe world better. For a company to know that
phenfen was dangerous inEurope and yet bring that drug over
into the USA to sell as a dietdrug, even knowing that it had
a past history of dangerousness, tellsme that Wyeth entire
corporate upper management needs to be ?ed. AndI do not want
to own a drug company of that sort of history of
placingmoney-grubbing over that of good science.Wyeth should
emulate Johnson & Johnson or Merck as companies that getthe
science before they ever think about money.I am wanting and
trying to get the PAF portfolio positioned for theyear 2004
such that it follows the VonNeumann Game Theory of theOptimal
Strategy for Playing the StockMarket. The key and centraltheme
of the OS of Stockmarket is switching campaigns of
theCrossover.Through these many years of playing the
stockmarket since 1978, Ibeleive I have found the OS of
StockMarket and am trying to arrangethe portfolio so that I
can faithfully abide by this OS for 2004 andmany years
beyond.Archimedes Plutoniumwhole entire Universe is just one
big atom where dotsof the electron-dot-cloud are
galaxies
===
Here I thought in this GOOGLE site, of whatever
God was restrictedand/or moderated to NASA/NSA/DoD, now
DHS.Seems there's some folks that certainly talk and act the
part likeGod, as they certainly are those willing and able to
terminatewhatever life as we know it, especially mine. Though
I guess I'm justhappy that I'm not one of those nice 
Cathars
having to dodge anotherround of exterminations by the Pope,
though our resident warlord hascertainly been doing his fair
share of mastering carnage based uponlies similar to but not
even nearly as good as what the Catholicchurch used to
justify their actions. Although, we've got the 
sowhat's the
difference as our ultimate quali?r that trumps
anythingPope.BTW; did anyone here bother to check out David
Sereda, and of hishttp://www.ufonasa.com/ EVIDENCE the case
for NASA UFOsIn some of my recent additions to the MAZDA like
Internal RocketRotary Combustion Engine (IRRCE sfc =
15+KW/kg), there seems we alsohave ourselves a wee bit of
lunar He3 to burn
off.http://guthvenus.tripod.com/gv-h2o2-irrce.htmThe taking
of lunar He3 is ?st come ?st served, that could bethe likes
of China or Russia because, we're obviously not
smartenough.Venus, not Mars, still offers life; via moon He3
could help turn thetrickI've got a few more words of wisdom
to offer on behalf of the ARTEMISPROJECT (lunar He3)
http://guthvenus.tripod.com/gv-lse-he3.htmThe rest of this
report isn't entirely related to energy so much as itrelates
to truth or consequences. Such as for this next topic of
therebeing other life NOT as we know it on Venus, that's
obviously opposingthe sorts of anti-humanity folks that
couldn't care less if our entireworld was destroyed by their
resident warlord. Perhaps these folks canget their next level
of future funding from the same source as Bush,Salem Laden.As
for making policy look like happenstance, and/or vice versa,
is keyto snookering folks.
http://guthvenus.tripod.com/moon-04.htmThough as for we
humans need not, and perhaps we should not ventureourselves
much beyond Venus L2 (VL2). Wouldn't want to contaminate
aperfectly good planet with our inferior DNA nor lack of
morals,especially of this group that's bashing honest
research just out ofspite. Besides, their stealth
donkey-carts could be far more lethalthan what our WMD
donkey-carts can manage.As far as our human physiology being
adaptable to pressure. Under suchpressure things are not
nearly as hot as we've been told, and you wontneed but a
fraction of a percent of O2. Of course, that degree
ofadaptation might have to be accommodated at a modus rate of
a few barsper day.http://guthvenus.tripod.com/venus-air.htmI
have a few other recent/ongoing comments on H2O2/C12H26 and
of
He3:http://guthvenus.tripod.com/gv-irrce.htmhttp://
guthvenus.tripod.com/gv-hybrid-irc.htmhttp://
guthvenus.tripod.com/gv-cm-ccm-01.htmhttp://
guthvenus.tripod.com/gv-lm-1.htmhttp://guthvenus.tripod.com/
radio-maybe.htm
===
Dear all,I want to ask what is the inverse
operation of Kroneck product?More speci?ally, we know that
matrix operationA*X*B=kron(A, B')*vec(X)where kron is the
Kronecker product of matrices as de?ed in matlab; vec(X)is
the stacked vector version of matrix X. All matrices are
square...Now I want to reverse the operation, suppose I have
a big matrix C,how to ?d A and B to get C=kron(A, B')?Under
what condition these A and B cannot be found? Then how to ?d
themapproximately, i.e., optimal in the mean-square sense or
under othercriteria? That's to say, ?d A and B, such that
kron(A, B')=C1 where C1 isa reasonably good approximation to
C?-Walala
===
walala:Maybe if you gave the homework questions
verbatim from the Prof, we could beof more help. It seems
your interpretation of your tasks is confusing somepeople.
Then again, if you really knew the questions, you wouldn't
beposting for answers.Jim> Dear all, I want to ask what is
the inverse operation of Kroneck product? More speci?ally,
we know that matrix operation A*X*B=kron(A, B')*vec(X) where
kron is the Kronecker product of matrices as de?ed in
matlab;vec(X)> is the stacked vector version of matrix X. All
matrices are square... Now I want to reverse the operation,
suppose I have a big matrix C, how to ?d A and B to get
C=kron(A, B')? Under what condition these A and B cannot be
found? Then how to ?d them> approximately, i.e., optimal in
the mean-square sense or under other> criteria? That's to
say, ?d A and B, such that kron(A, B')=C1 where C1is> a
reasonably good approximation to C?> -Walala
===
> walala:
Maybe if you gave the homework questions verbatim from the
Prof, we couldbe> of more help. It seems your interpretation
of your tasks is confusing some> people. Then again, if you
really knew the questions, you wouldn't be> posting for
answers. Jim > Dear all,> I want to ask what is the inverse
operation of Kroneck product?> More speci?ally, we know
that matrix operation> A*X*B=kron(A, B')*vec(X)> where
kron is the Kronecker product of matrices as de?ed in
matlab;> vec(X)> is the stacked vector version of matrix X.
All matrices are square...> Now I want to reverse the
operation, suppose I have a big matrix C,> how to ?d A and
B to get C=kron(A, B')?> Under what condition these A and B
cannot be found? Then how to ?dthem> approximately, i.e.,
optimal in the mean-square sense or under other> criteria?
That's to say, ?d A and B, such that kron(A, B')=C1 
where
C1> is> a reasonably good approximation to C?> -Walala>Dear
Jim,This really isn't my professor's homework 
problem... our
school is now inbreak so there are no damn homeworks right
now...This is a problem I am currently interested in...
please tell me which partof my description confuses you? Then
I can make myself clearer...Rgs,-Walala
===
How would I compute
the probability of a draw in tac tac toe (noughts and crosses)
assuming each player was playing randomly (ie no thought
involved). What about the probability that player 1
wins?--Interbangchange nospam to optimusprime to
reply
===
>How would I compute the probability of a draw in
tac tac toe (noughts >and crosses) assuming each player was
playing randomly (ie no thought >involved). What about the
probability that player 1 wins?Assuming X goes ?st and that
the players alternate until the grid isfull, the number of
?al complete grids can be found by computing 9C4- this is
the number of ways of choosing the 4 cells that get OsOf
course the 9C4 includes many games that are merely rotations
orre?ns of each other. It also ignores the fact that
many of thegames would really be over before the grid is
full. But you may ?dthat ignoring such issues provides the
most straightforward approach.In particular, all the
positions among the 9C4 can be regarded asequally likely,
which will not be the case if you consider positionsat the
real end of the game.What you have to do next is ?d the
number of drawn positions, i.e.the number of ?led-in grids
which do not contain a winning lineeither for X or for O. You
could approach this by enumerating theplaces in the grid where
a winning line can arise and then consideringthe possible
arrangements of the other symbols. You will need to bevery
careful to avoid counting twice the positions with more than
onewinning line.Hope this helps
===
> > So
you know that tan' pi/4 = whatever.> Now what does that
*mean*?> (Can you recall the *de?ition* of
derivative?)> It means that you can apply 
l'Hopital's
rule without knowing in> advance what lim_{x -> pi/4}
(tan(x) - 1)/(x - pi/4) is.> Try again! That's no
de?ition of derivative.> How do you use L'H's 
rule to
calculate this limit without> already knowing what it is?>
> The de?ition of derivative of tan(x) may be deduced from
the> derivatives of sin(x) and cos(x) and the quotient rule
for> derivatives, the derivative of the numerator and
denominator of> (tan(x) - 1)/(x - pi/4) can be found without
being aware that> it is a difference quotient.> If f(x) =
tan(x) - pi/4 = sin(x)/cos(x), then> f'(x) = (cos(x)*cos(x) -
sin(x)(-sin(x))/cos(x)^2 + 0> = 1/cos(x)^2> Excellent:
you can compute derivatives. Alas you seem> to have
forgotten what they are. :-(> You have (d/dx)(tan x -
pi/4) = 1/cos^2 x [dunno what the pi/4 is doing,> but it's
irrelevant anyway]. Now what does that mean?> (Heavy
hint: apply the de?ition of derivative.)> and if g(x) =
x -pi/4 then g'(x) = 1> Any point to this?> Since
f(pi/4) = g(pi/4) = 0 but g'(pi/4) and f'(pi/4) are>
continuous and nonzero at x = pi/4, L'Hopital applies.>
But why waste time using L'H?It is not a matter of whether it
is optimal to do it the way I> showed, but merely a question
of whether it is possible, and I have> shown it to be
possible.No you didn't --- you used the limit itself in the
processof applying L'H :-(-- Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the
last laugh. Alan Partridge, _Bouncing Back_ (14 times)
===
>
[snip]>In his Commutative Algebra, D. Eisenbud has another
de?ition:>De?ition 2. Let R be a commutative ring. An
element p of R is prime if> p is not a unit and the
following is true: If a and b are elements of>R such that p
divides ab, then p divides a or b.>Is De?ition 2 common
in the commutative algebra community?> It is common in
algebraic number theory as well.Then I must wonder what the
common de?ition of divides is. According> to the de?ition
which I typically use, 0 divides 0. And then according> to
De?ition 2, we would have 0 being a prime, which surely we
don't> want.Not necessarily --- the zero ideal isn't 
prime in
every ring.Def 2 just says that p is prime iff p generates a
prime ideal.Of course, in algebra, it's prime ideals, not
prime elements, thatare of major interest.-- Robin Chapman,
www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to say, I had the
last laugh. Alan Partridge, _Bouncing Back_ (14
times)
===
bccDear Gary, Jack and Tony :worked for some time
on the notion of Self ReferentialNoise as models of reality.
Wheelers's spacetime FOAMis just another interpretation of
Noise. You couldtake that Noise as your absolute frame
ofreference.JS: The new EINSTEIN results seem to argue
against this model?I am not sure of that of course. Sirag
says the foam is not reallyrandomly chaotic but coherently
harmonic and that sets thenumber of extra dimensions in
Calabi-Yau space? On the other handChristian Beck seems to
agree with you and says he can determinethe 25 epicycles of
the standard model (some say only 17) fromchaotic strings.
This would seem choose a de?ite location on Susskind's
Landscape in opposition to the WAP ideas of chaotic papers
and books by Lee Smolin.CC: One day you may want to look at
:Carlos Castro `` The String Uncertainty RelationsFollow from
the New Relativity Principle .Foundations of Physics. {bf 8} (
2000 ) page 1301.for a way to derive the stringy uncertainty
relationsfrom ?st principles.JS: Is the claim being made
that the new term beyond Heisenberg uncertainty is the source
of irreversibility as in the arrow of time? Remember Hawking
talks about a new source of uncertainty although Susskind
seems to think that is wrong?Note also Ed Witten's formula
generalizing Heisenberg's quantum uncertainty principle, i.e.
eq. (5.9) p. 136Delta X > h/DeltaP + alpha'(DeltaP)/hThe
second gravity-string source of uncertainty should give the
irreversible statistical arrow of time not found when alpha'
= 0, i.e. in?ite string tension, or in?ite space-time
stiffness of action without reaction as is also found in the
signal locality of orthodox quantum theory in sense of Antony
Valentini's papers.Note the conformal lookw = 1/z +
alpha'z
===
> proofs> why do authors usually consider the
proofs so important in mathematical> texts ?Because that is
what mathematics is without.> Isn't it usually better when
only a few people read and check the proofs,No. If you refuse
study proofs then you are cravenly deferring to authority.--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.htmlNeedless to
say, I had the last laugh. Alan Partridge, _Bouncing Back_
(14 times)
===
> proofs> why do authors usually
consider the proofs so important in mathematical> texts
?Because that is what mathematics is without.Is this sentence
a word without or something?> Isn't it usually better when
only a few people read and check the proofs,No. If you refuse
study proofs then you are cravenly deferring to
authority.************************David C. Ullrich
===
 : >
----------------------------- <^> <(?e? : <^>
----------------------------- : > : > : > Any number having
this form can be produced by an RM. : > Some examples of
computable numbers: : > : > (0) : > (1) : > 11(0) : > 01(110)
: > : > The length of the initial string plus the length of
the repeating string : > must be less than or equal to the
number of states of the RM that : > produced the number. : >
: > : > I don't see this. Here is a simpli?d conceptual
diagram of a Russell : Machine. : > : > s1 s2 s6 : > 0 s5 : >
--------------------- : > 1 s3 : > s4 s7 : > s8 : > : > There
is one piece of information missing, which state s8 goes to.
: > Add to this s8 -> s7 : > : > This RM will output 0 0 1 1
0 0 1 1 1 1 1 1 1 1 : > notation is 0 0 1 1 0 0 (1 1) : > : >
If it can't halt then the states get reused. : : I assume that
s1->s2->s3->s4->s5->s6->s7->s8->s8. : The ouput is
001100(1).Then it DOES halt, DUMBASS. s8 *IS* a halt
state.Are you now going to retract your claim that RMs don't
halt becausethey don't have halt states?
===
 : > What kinds
of numbers are we talking about here? : > Rational? Real?
Complex? Supernatural? In?itesimal? : > : I have made the
proof a little more rigorous.But you haven't answered the
question.WHAT IS a number??? : First, I de?e how to compute
a computable number.That is ENTIRELY premature. FIRST, you
have to de?e NUMBER! : I do this with a very non-standard
type of Turing machine. : Let's call these Russell machines
(RM). : A number is computable if it can be represented by :
the in?ite binary string produced by a RM.This is just
stupid. A TM can do anything you need to do. : A RM has a
single, in?itely long tape. : Every position of the tape is
initially set to 0. : A RM starts at the leftmost end of the
tape. : RMs can perform two operations, labeled 0 and 1. : :
0 = move write head one position to the right : 1 = write a 1
to current location and move one position right : : RMs never
read from the tape because they would always read 0. : RMs
never halt because they have no halt states.The non-halting
is stupid. It basically makes it impossible tohave a
well-de?ed result. If you ever DO have a well-de?edresult,
observers can always back-allege that the machine actuallyDID
halt, as soon as the result BECAME well-de?ed. You can
allegethat it is still running but WE can allege that what it
is now doingno longer deserves to be CALLED running. : The
state of an n-state RM can be speci?d with : (1 + Ceil(
log2( n ) ) ) bits where the ?st bit represents the :
operation to perform and the next Ceil(log(2(n)) bits :
represent which state to switch to.This is ungrammatical. If
the bits represent which state to switchto then they DO NOT
represent the state you are CURRENTLY in!But you SAID the
state of an n-state RM can be speci?d with....I guess
speci?d is just ambiguous. : An n-state RM can be fully
speci?d by listing all states : which requires n * (1 +
Ceil(log2(n))) bits.OK. : Some examples of RMs and their
output: : : States RM Output : 1 0 (0)... : 1 1 (1)... : 2 01
11 01(1)... : 2 11 00 (10).... : 3 001 010 100 (001)... : :
The output of a RM has the following form: : There is a
?ite, possibly empty initial string that is written once, :
and a ?ite, never empty repeating string that is written
over and over. : A computable number can be written as the
initial string followed by : the repeating string in
parenthesis.In other words, this thing only produces rational
numbers.That is a stupid limitation. : Any number having this
form can be produced by an RM. : Some examples of computable
numbers: : : (0) : (1) : 11(0) : 01(110) : : The length of
the initial string plus the length of the repeating string :
must be less than or equal to the number of states of the RM
that : produced the number. : : Now, prove that no set can
contain every RM computable number.But it can; this is just
the rational numbers. They are a set. : The standard diagonal
argument has a problem - it is usually : impossible to
determine if the diagonal number produced is : computable.No,
it isn't; that is NOT the problem.I just threw the rest of it
away.Bothering to create an alternative paradigm IS
STUPID.You should have STUCK to TMs. You relieve yourself of
theburden of explanation to everybody (of what
non-standardthing you mean) -- everybody ALREADY knows what a
TM is.And whenEVER you are asked a direct question,JUST ANSWER
IT, if you are going to respond.WHAT KIND OF *NUMBERS* are you
talking about (forjudging computability)? Natural? Real?
Complex?Rational? Supernatural? In?itesimal? *ANSWER* THE
ING *QUESTION*, DUMBASS!
===
> : Let S be the set of all
computable numbers> : and assume S is countable.I have made
the proof a little more rigorous.First, I de?e how to
compute a computable number.> I do this with a very
non-standard type of Turing machine.> Let's call these
Russell machines (RM). > A number is computable if it can be
represented by> the in?ite binary string produced by a RM.
[...]Let C be the set of all RM-computable binary strings,
according to your de?ition. To show that C is countable, it
suf?es to demonstrate a one-to-one function f : C -> N,
where N is the set of natural numbers { 0, 1, 2, ... }. This
is not dif?ult.First, let's agree that any RM M with n
states can be uniquely encoded as a natural number. How we
choose to do so isn't important, as long as we agree upon the
rule -- so let's say for argument's sake that we use 
your idea
as a basis. To encode M, write 1, followed by n zeroes,
followed by another 1, and then a sequence of n strings with
ceil(lg(n)) + 1 bits each, encoding the transition rules in
ascending numerical order as you described. Treat the
resulting string as a natural number written in binary, and
let that be the encoding for M. Whatever this number is, I'll
denote it E(M). It should be obvious that if M1 and M2 are
distinct RM's, then E(M1) <> E(M2).Now for any RM-computable
string c., let RM(c) be the set of all RM's that compute c.
There must be AT LEAST one member of this set, since c is
RM-computable*. Let M(c) be the unique element of RM(c) that
ful?s these two properties: 1. M(c) in RM(c) 2. For all M in
RM(c), E(M(c)) <= E(M).In other words, M(c) is the RM for c
that maps to the smallest natural number among all the RM's
that compute c, under the RM-encoding rule. It's not
important that we pick this one in particular, but doing so
eliminates all ambiguity.So: We can now de?e a one-to-one
function f : C -> N as follows: f(c) := E(M(c))Quickly, let's
argue that this is one-to-one: Let c1 and c2 be RM-computable
strings with c1 <> c2. Then M(c1) <> M(c2)**, and by the
de?ition of E, E(M(c1)) <> E(M(c2)), so we're all set.This
suf?es to show that the false diagonalization proof you
sketched out is inapplicable here.-M* And, actually, there
are in?itely many RM's for any RM-computable string. The
proof is left as an exercise for the reader. ;)** I hold this
truth to be self-evident. But if you disagree, it's not so
terrible to prove it from your RM de?itions.P.S.- I mean for
<> to denote the does not equal relation.P.P.S.- This level of
detail is overkill for just getting the intuition, but
imprecision can be even MORE confusing, I ?d.--
http://www.dartmouth.edu/~sting/ | Dartmouth College,
Hanover, NH, USA
===
in message > : Let S be the set of all
computable numbers> : and assume S is countable.> I have
made the proof a little more rigorous.> First, I de?e how
to compute a computable number.> I do this with a very
non-standard type of Turing machine.> Let's call these
Russell machines (RM).> A number is computable if it can be
represented by> the in?ite binary string produced by a RM.
[...] Let C be the set of all RM-computable binary strings,
according to your> de?ition. To show that C is countable, it
suf?es to demonstrate a> one-to-one function f : C -> N,
where N is the set of natural numbers {> 0, 1, 2, ... }. This
is not dif?ult.I didn't think so until a few days 
ago.It's
harder than it looks.> So: We can now de?e a one-to-one
function f : C -> N as follows: f(c) := E(M(c)) Quickly,
let's argue that this is one-to-one: Let c1 and c2 be>
RM-computable strings with c1 <> c2. Then M(c1) <> M(c2)**,
and by the> de?ition of E, E(M(c1)) <> E(M(c2)), so we're
all set.Let x be the string produced by the comparator.By
construction, x was not produced by f().E() depends on f().
E(M(x)) is unde?ed.x represents a RM computable natural
number.Russell- 2 many 2 count
===
 > Let C be the set of all
RM-computable binary strings, according to your> de?ition.
To show that C is countable, it suf?es to demonstrate a>
one-to-one function f : C -> N, where N is the set of natural
numbers {> 0, 1, 2, ... }. This is not dif?ult.I didn't think
so until a few days ago.> It's harder than it looks.So: We can
now de?e a one-to-one function f : C -> N as follows:> f(c)
:= E(M(c))> Quickly, let's argue that this is one-to-one:
Let c1 and c2 be> RM-computable strings with c1 <> c2. Then
M(c1) <> M(c2)**, and by the> de?ition of E, E(M(c1)) <>
E(M(c2)), so we're all set.Let x be the string produced by
the comparator.I'm sorry, perhaps I missed something. What
comparator?> By construction, x was not produced by f().> E()
depends on f(). E(M(x)) is unde?ed.> x represents a RM
computable natural number.Setting aside for the moment that I
do not know what string you mean by x, I can safely assert
that E does not depend upon f. I chose E to be a simple
encoding rule, so that we could be easily convinced that it
is a one-to-one mapping from RM's to natural numbers. Its
de?ition does not depend upon f's de?ition.What do you mean
by an RM-computable natural number? Your de?ition of
RM-computability talks about binary strings of in?ite
length. Due ot their regular structure, you can ENCODE such a
string uniquely as a natural number. That's essentially what I
proved above. But you did not show a one-to-one mapping from
RM-computable strings to natural numbers in your original
de?ition! The obvious mapping -- just dropping the
parentheses and gluing the bits together) doesn't quite work,
because (for example) all these distinct RM-computable strings
map to the same number: 010(10), 0010(10), 00010(10), ...If
you don't like referring to a speci? RM, an alternate proof
formulation would be to ?st prove that RM-computable strings
have the structure you described: wx+ where w in (0+1)* and x
in (0+1)(0+1)*. Once you prove that, you can encode any
RM-computable string in binary by mapping 0 => 0, 1 => 11,
and . to 10 and writing .w.x (the dots guarantee a unique
natural number interpretation even when w begins with one or
more zeroes). Again, the choice of code is unimportant except
insofar as we have to agree what it is.-M--
http://www.dartmouth.edu/~sting/ | Dartmouth College,
Hanover, NH, USA
===
 Let x initally be (0).> If s has the
form 111...111(0) and the length of the initial> segment of s
is longer or equal to the length of the initial> segment of x
then take the initial segment of s, append a 1, and> make
this new string the initial segment of x. Examples: s x> (0)
1(0)> 1(0) 11(0)> 11(0) 111(0) Prove that x differs from
every member of S. x differs from every s that does not end
with (0).> x differs from every s with an inital segment not
of the form> 111.111. x differs from every s with form
111...111(0).> Therefore, x differs from every member of S.>
Well ... let S be the set of al RM computable numbers of the
form> 1*(0) (any number of sequential ones followed by a
repeating of> zeros). In this way, you will not be able to
compute a x which> differs from every possible member of S
(at least not using your> technique).> Why not?> x will
differ from every member of S and x will be a RM computable>
number.> Another way to look at his :> if all the numbers in
a set S or of the form 111...11(0), you can de?ea> set S'
(containing natural numbers) such that an element of S which
has a> sequence of m 1's corresponds to an element m of 
S'.
Your x' would thenbe> m+1. Now if you consider the set S I
used in my previous post, it would> correspond tot the
entire set of natural numbers. As for each m,contained> in
this set, m+1 is also contained, there cannot exist an x'
which doesnot> belong to this set.Earlier in this post,
someone said the natural numbers are computable.Let's use
your set as an example. We can interpret the strings right
toleft.Now, each binary string represents a natural number in
base 1.Let's call the set S.S(0) = (0)S(1) = (0)1S(2) =
(0)11...I give a constructive method of creating a RM
computable number, x,that is not in S. How can my method fail
to produce x?You assume that every string that corresponds to
a ?ite number is in S.This is not true if my proof is
correct.I can only think of two reasons why I would not be
able to compute x.1) S contains a member with in?itely many
1's2) S contains a member with so many 1's that 
addingone
more 1 results in an in?itely long string of 1's.If you can
think of another reason why x can not be computed,please post
it.Can you show that x is impossible to compute?Russell- 2
many 2 count
===
> Let x initally be (0).> If s has the
form 111...111(0) and the length of the initial> segment of
s is longer or equal to the length of the initial> segment
of x then take the initial segment of s, append a 1, and>
make this new string the initial segment of x.> Examples:>
s x> (0) 1(0)> 1(0) 11(0)> 11(0) 111(0)> Prove that x
differs from every member of S.> x differs from every s that
does not end with (0).> x differs from every s with an inital
segment not of the form> 111.111. x differs from every s with
form 111...111(0).> Therefore, x differs from every member of
S.> Well ... let S be the set of al RM computable numbers
of the form> 1*(0) (any number of sequential ones followed
by a repeating of> zeros). In this way, you will not be
able to compute a x which> differs from every possible
member of S (at least not using your> technique).> Why
not?> x will differ from every member of S and x will be a
RM computable> number.> Another way to look at his :> if
all the numbers in a set S or of the form 111...11(0), you
can> de?e > a> set S' (containing natural numbers) such
that an element of S which> has a sequence of m 1's
corresponds to an element m of S'. Your x'> would 
then >
be> m+1. Now if you consider the set S I used in my
previous post, it> would correspond tot the entire set of
natural numbers. As for each> m, > contained> in this
set, m+1 is also contained, there cannot exist an x' which>
does > not> belong to this set.Earlier in this post, someone
said the natural numbers are computable.> Let's use your set
as an example. We can interpret the strings right> to left.>
Now, each binary string represents a natural number in base
1.> Let's call the set S.S(0) = (0)> S(1) = (0)1> S(2) =
(0)11> ...I give a constructive method of creating a RM
computable number, x,> that is not in S. How can my method
fail to produce x?Because every x would be of the form
(0)11..1, and because of the de?ition of S, any such RM
number is contained in S.> You assume that every string that
corresponds to a ?ite number is in> S. This is not true if
my proof is correct.That was how S was de?ed : it contains
all the strings of the form (0)11..1 (this can be considered
as all unary representations of natural numbers).> I can only
think of two reasons why I would not be able to compute x.1) S
contains a member with in?itely many 1's> 2) S contains a
member with so many 1's that adding> one more 1 results in an
in?itely long string of 1's.If you can think of another
reason why x can not be computed,> please post it.3) for any
s in S, s-with-a-1-added-to-the-back is also contained in
SCan you show that x is impossible to compute?As there can
not exist an element which is both member of a set and not
contained in the set, yes.Russell> - 2 many 2 count> --
PentoDe wereld was soep, en het denken meestal een vork,tot
smakelijk eten leidde dat zelden. - H. Mulisch
===
> I give
a constructive method of creating a RM computable number, x,>
that is not in S. How can my method fail to produce x? Because
every x would be of the form (0)11..1, and because of the>
de?ition of S, any such RM number is contained in S.The
de?ition of S leads to contradiction.S can not contain every
natural number.> You assume that every string that corresponds
to a ?ite number is in> S. This is not true if my proof is
correct. That was how S was de?ed : it contains all the
strings of the form> (0)11..1 (this can be considered as all
unary representations of natural> numbers).I show that there
exists an RM computable number of the form (0)11...1that is
not in set S. I can even show how this number can be
constructedby a Turing machine.RM's are a subset of TMs.Any
RM can be emulated by a universal Turing machine (UTM).We are
only concerned with the subset of RM's that output an
initial,?ite and contiguous string of 1's followed by an
in?ite string of 0's.The UTM can generate the output tape
for each of these RM's.These tapes are then read by a second
TM I will call a Comparator (CTM).The CTM compares each input
tape with the CTM's output tape.If the input tape has a longer
initial string of 1's, the CTM rewinds and1.After all of the
RM tapes have beed read by the CTM we examine theoutput of
the CTM. This tape must contain the representation of
anatural number.Every tape read by the CTM represents a
natural number.The CTM output tape contains the
representation of the successorof some member of S. The
successor of a natural number is anatural number.S can not
contain a representation of every natural number.> I can only
think of two reasons why I would not be able to compute x.>
1) S contains a member with in?itely many 1's> 2) S contains
a member with so many 1's that adding> one more 1 results in
an in?itely long string of 1's.> If you can think of
another reason why x can not be computed,> please post it. 3)
for any s in S, s-with-a-1-added-to-the-back is also contained
in SNot true. I show how to construct such a number that is
not in S.> Can you show that x is impossible to compute? As
there can not exist an element which is both member of a set
and not> contained in the set, yes.The de?ition of S leads
to contradiction.S can not exist.Russell- 2 many 2 count
===
>
> I give a constructive method of creating a RM computable
number, x,> that is not in S. How can my method fail to
produce x?You have yet to say What KIND of numbers RM numbers
CAN OR CAN'T be.Are they natural? Rational? Real? Complex?
Supernatural? In?itesimal?IT MATTERS. ANSWER the question!>
> Because every x would be of the form (0)11..1, and because
of the> de?ition of S, any such RM number is contained in
S.The de?ition of S leads to contradiction.No, it doesn't.>
S can not contain every natural number.It can and it does.YOU
de?ed it. Do you have any idea howSTUPID this makes you look,
de?ing a set and thensaying that it can't contain things it
obviously contains?> You assume that every string that
corresponds to a ?ite number is in> S. No, we don't assume
it, we just NOTICE it.You de?ed S as the class of all outputs
of RMs that printedsome natural number of 1s and then
stopped.> This is not true if my proof is correct.Ergo, your
proof is not one.> That was how S was de?ed : it contains
all the strings of the form> (0)11..1 (this can be considered
as all unary representations of natural> numbers).YOU de?ed S
that way. So S contains all natural numbers.> I show that
there exists an RM computable number of the form (0)11...1>
that is not in set S. No, you don't.> I can even show how
this number can be constructed> by a Turing machine.No, you
can't. > RM's are a subset of TMs.No, 
they're not. For one
thing, you alleged that they never halted.Are you going to
retract that?> Any RM can be emulated by a universal Turing
machine (UTM).> We are only concerned with the subset of RM's
that output an initial,> ?ite and contiguous string of 1's
followed by an in?ite string of 0's.Since RM output tapes
were supposed to START OUT as all 0's, can't thethe RM 
just
HALT after it gets through printing all its 1's?> The UTM can
generate the output tape for each of these RM's.And put it
where, exactly? TM's, by default, only have ONE output tape.>
These tapes are then read by a second TM I will call a
Comparator (CTM).No, they aren't. There are an in?ite number
of these tapes and the CTMnever ?ishes reading them all. >
The CTM compares each input tape with the CTM's output
tape.The CTM does NOT HAVE an output tape, other than its
input tape.That's just how TMs ARE DEFINED.
===
----------------------------- <^> <(?)> <^>
-----------------------------> RM's are a subset of TMs.> Any
RM can be emulated by a universal Turing machine (UTM).> We
are only concerned with the subset of RM's that output an
initial,> ?ite and contiguous string of 1's followed by an
in?ite string of 0's.>So you're not proving there is 
no
complete RM list, 1st you areproving there is no complete
restricted RM list?Is this based on modelling the diagonal
number from your previous post?<quote > and then prove that
it is not in my list.> 0.000...> 0.1000...> 0.11000...>
0.111000...> 0.1111000...> 0.11111000...> 0.111111000...>
...> Good luck trying to prove the diagonal number is> not
in the list using a countable number of operations. Seems too
easy, surely? If I understand you correctly the diagonal>
number is: 0.11111.... Every number in your list consists of
a ?ite string of 1s followed> by zeros (000...). The
diagonal number doesn't. Ergo it isn't in the> list.I 
can ?d
an entry in my list that matches the diagonal number to
any?ite number, n, of positions. This is true for all n.My
list is in?ite.There is at least one entry in my list that
has a 1 in every ?iteposition.How can you prove the diagonal
number is not in my list?Russell</quote><quote>convenience,
I've changed the layout very slightly... > 0.111000... >
0.1111000... > 0.11111000...> 0.111111000...
^--------</quote>Herc</bah>
===
 -----------------------------
<^> <(?)><^> -----------------------------> RM's are a
subset of TMs.> Any RM can be emulated by a universal Turing
machine (UTM).> We are only concerned with the subset of RM's
that output an initial,> ?ite and contiguous string of 1's
followed by an in?ite string of0's.So you're not 
proving
there is no complete RM list, 1st you are> proving there is
no complete restricted RM list?This is the only subset of
RM's anyone has concerns about.I think everyone agrees my
method works for all the other RMs.> Is this based on
modelling the diagonal number from your previous
post?Partly.> <quote> and then prove that it is not in my
list.> 0.000...> 0.1000...> 0.11000...> 0.111000...>
> 0.1111000...> 0.11111000...> 0.111111000...> ...>
Good luck trying to prove the diagonal number is> not in
the list using a countable number of operations.> Seems too
easy, surely? If I understand you correctly the diagonal>
number is:> 0.11111....The diagonal method can be converted
into the computable number proof.Of course, when I responded
to your thread I was arguing against Cantor'sdiagonal proof.
I am arguing for it in this thread.Forcing the diagonal
number to be computable shows that invokingin?ity is like
killing a ?h a machine gun.All we really need to prove
it that the diagonal has more 1's thanany of the real numbers
in the list above. The diagonal can havea ?ite number of 1's
and still not be in the list.I suspect that the diagonal
proof can be modi?d to show thatthe rational numbers are
uncountable.Let R be the set of all rational numbers, r, such
that 0 <= r < 1.If R can be ordered such that the diagonal is
a rational number,the diagonal proof shows the rationals to
be uncountable.Most people will argue that the diagonal of
this set must bean irrational number. I have never seen a
proof of this.It is easy to come up with a list of rationals
that have a rationaldiagonal. The set I give above is one
such set.Russell- 2 many 2 count
===
> Let R be the set of all
rational numbers, r, such that 0 <= r < 1.> If R can be
ordered such that the diagonal is a rational number,> the
diagonal proof shows the rationals to be uncountable.And if
my grandmother had wheels, she'd be a wagon.There are several
ways to put that subset of the rational numbers intoone-to-one
correspondence with the positive integers.> Most people will
argue that the diagonal of this set must be> an irrational
number. I have never seen a proof of this.> It is easy to
come up with a list of rationals that have a rational>
diagonal. The set I give above is one such set.Of course,
that set was not all of the rationals in that range.Here is
an example of the beginning of a complete
list:01/21/32/31/43/41/52/53/54/51/65/61/7...Identifying a
rational given its position (or vice-versa) is probablygoing
to require generating the list up to that position, but this
listis obviously complete. Therefore, the set of rational
numbers in [0,1)is countable.You want a proof that any such
complete list has an irrational diagonal?Okay:Assume that a
complete list with a rational diagonal can be found.
Thediagonalization process would generated a rational number
not in thelist. This would contradict the assumption.
Therefore, either the listis incomplete or the diagonal is
not rational. (My list is complete, sothe diagonal must not
be rational. Your list had a rational diagonal,but was
obviously incomplete.)When applied to an allegedly complete
list of the reals in [0,1), thediagonally-generated number
does not have the option of being other thanreal, so the only
possible conclusion is that the list was incomplete. -- Daniel
W.
Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes
/039 53 36 N / 086 11 55 W
===
 > Let R be the set of all
rational numbers, r, such that 0 <= r < 1.> If R can be
ordered such that the diagonal is a rational number,> the
diagonal proof shows the rationals to be uncountable. And if
my grandmother had wheels, she'd be a wagon. There are
several ways to put that subset of the rational numbers into>
one-to-one correspondence with the positive integers. > Most
people will argue that the diagonal of this set must be> an
irrational number. I have never seen a proof of this.> It is
easy to come up with a list of rationals that have a
rational> diagonal. The set I give above is one such set. Of
course, that set was not all of the rationals in that
range.Of course.> Here is an example of the beginning of a
complete list: 0> 1/2> 1/3> 2/3> 1/4> 3/4> 1/5> 2/5> 3/5>
4/5> 1/6> 5/6> 1/7> ... Identifying a rational given its
position (or vice-versa) is probably> going to require
generating the list up to that position, but this list> is
obviously complete. Therefore, the set of rational numbers in
[0,1)> is countable.Maybe. That is the point of the proof.>
You want a proof that any such complete list has an
irrational diagonal?> Okay: Assume that a complete list with
a rational diagonal can be found. The> diagonalization
process would generated a rational number not in the> list.
This would contradict the assumption. Therefore, either the
list> is incomplete or the diagonal is not rational. (My list
is complete, so> the diagonal must not be rational. Your list
had a rational diagonal,> but was obviously incomplete.)How
are you proving your list is complete?To prove that the set
contains all rational numbersyou will have to show there is
no way to orderthe set such that the diagonal is
rational.There are a lot of ways to order a set of
rationals.Russell- 2 many 2 count
===
> How are you proving
your list is complete?Any rational in [0,1) can be uniquely
written as a ratio between twocoprime nonnegative integers
m/n. That is an entry among the ?stn(n-1)/2 + 1 entries on
the list, although specifying the exact locationinvolves a
summation on the Euler phi-function. Anyway, that entry onthe
list corresponds to no other rational. (This last proviso
ismoderately important, because some people like to present
completelists of reals in which a given list entry can have
more than one realnumber associated with it.)> To prove that
the set contains all rational numbers> you will have to show
there is no way to order> the set such that the diagonal is
rational.I just proved that the set contains all rational
numbers in [0,1).To prove that a given integer is even, is it
necessary to show both thatits units digit in base two is 0
and that its units digit in base ten isin {0,2,4,6,8}?> There
are a lot of ways to order a set of rationals.If you are
talking about the complete set of rationals, the number
ofways is the same as the number of real numbers. If you
think a relevantordering exists, either point out a ? my
proof or describe theordering.Otherwise, you might as well
quibble with a proof that the sum of any?ite set if even
numbers is not odd by pointing out that there are alot of
ways to choose a set of even numbers.-- Daniel W.
Johnsonpanoptes@iquest.nethttp://members.iquest.net/~panoptes
/039 53 36 N / 086 11 55 W
<-sOdnVAq4PkNeXGiRVn-sA@comcast.com>
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<1g6p2lp.1nq0zj71e3df92N%panoptes@iquest.net>
<9tCdnYk507_tD3KiRVn-hQ@comcast.com>
===
 > Let R be the set
of all rational numbers, r, such that 0 <= r < 1.> If R can
be ordered such that the diagonal is a rational number,> the
diagonal proof shows the rationals to be uncountable.> And
if my grandmother had wheels, she'd be a wagon.> There are
several ways to put that subset of the rational numbers into>
one-to-one correspondence with the positive integers.> Here is
an example of the beginning of a complete list:> 0> 1/2>
1/3> 2/3> 1/4> 3/4> 1/5> ...> Identifying a rational given
its position (or vice-versa) is probably> going to require
generating the list up to that position, but this list> is
obviously complete. Therefore, the set of rational numbers in
[0,1)> is countable. Maybe. That is the point of the proof.et
cetera, et cetera. It's equivalent to the abbreviation
Q.E.D.There's no reason to say Maybe. [I repeat my opinion
that you should search Google before continuingto post in
this thread. Pedagogy is fun, and I'll be the ?st to admitI
have bene?ted greatly from it, but it tends to clutter up
the groupsif it's let run amok, IMHO.]> How are you proving
your list is complete? Actually, he didn't -- he just said,
It's obvious, and moved on.A rigorous proof would use the
traditional diagonal method -- NOTCantor's diagonalization,
but a different kind of diagonal analogyI think attributed to
Godel. It involves laying out the plane of(positive) rational
numbers as follows: 1/1--1/2 1/3--1/4 ... .' .' 
.' 2/1 2/2
2/3 2/4 ... | .' .' 3/1 3/2 3/3 3/4 ... .' 
4/1 4/2 4/3 4/4
... | ... ...You'll need a ?ed-width font to see the
ASCII-art line thatI've drawn zigzagging across the diagonals
of the grid. This lineobviously passes through every number in
the grid. And everypositive rational number is *somewhere* on
that grid, as you cansee from the way it's laid out. So the
line passes through everypositive rational number, in
sequence. Straighten out theline, and remove duplicates, and
add zero and the negative rationals,and you're done -- you
have a complete list of all the rationals.It begins 0, 1, -1,
1/2, -1/2, 2, -2, 3, -3, 1/3, -1/3, 1/4, -1/4, 2/3, ...and
continues to in?ity. Now match up the elements of thatlist,
one-to-one, with the positive integers 1, 2, 3, 4, 5, 6, 7,
8, 9, 10, 11, 12, 13, 14, ...Ta-da! [The next bit has been
re-ordered for clarity.]> To prove that the set contains all
rational numbers> you will have to show there is no way to
order> the set such that the diagonal is rational.> There are
a lot of ways to order a set of rationals. > You want a proof
that any such complete list has an irrational diagonal?>
Okay:> Assume that a complete list with a rational diagonal
can be found. The> diagonalization process would generated a
rational number not in the> list. This would contradict the
assumption. Therefore, either the list> is incomplete or the
diagonal is not rational.[Q.E.D.]-Arthur
===
----------------------------- <^> <(?)> <^>
-----------------------------> RM's are a subset of TMs.>
Any RM can be emulated by a universal Turing machine (UTM).>
We are only concerned with the subset of RM's that output an
initial,> ?ite and contiguous string of 1's followed by an
in?ite string of> 0's.> So you're not proving there 
is no
complete RM list, 1st you are> proving there is no complete
restricted RM list? This is the only subset of RM's anyone
has concerns about.> I think everyone agrees my method works
for all the other RMs. > Is this based on modelling the
diagonal number from your previous post? Partly. > <quote >
and then prove that it is not in my list.> 0.000...>
0.1000...> 0.11000...> 0.111000...> 0.1111000...>
0.11111000...> 0.111111000...> ...> Good luck
trying to prove the diagonal number is> not in the list
using a countable number of operations.> Seems too easy,
surely? If I understand you correctly the diagonal> number
is:> 0.11111.... The diagonal method can be converted into
the computable number proof. Of course, when I responded to
your thread I was arguing against Cantor's> diagonal proof. I
am arguing for it in this thread. Forcing the diagonal number
to be computable shows that invoking> in?ity is like killing
a ?h a machine gun. All we really need to prove it that
the diagonal has more 1's than> any of the real numbers in
the list above. The diagonal can have> a ?ite number of 1's
and still not be in the list. I suspect that the diagonal
proof can be modi?d to show that> the rational numbers are
uncountable. Let R be the set of all rational numbers, r,
such that 0 <= r < 1.> If R can be ordered such that the
diagonal is a rational number,> the diagonal proof shows the
rationals to be uncountable. Most people will argue that the
diagonal of this set must be> an irrational number. I have
never seen a proof of this.This is my argument against the
diagonal. I can ?d a rationalon the list of computables that
equals any speci? representation (?ite)of the ?irrational'
diagonal construct number.If the rational number equals the
irrational number then its the same number.I'm pretty sure
Cantor falls on its head, I can give an algorithm for all
numbersUTM(Z), and increasing portions of the list it makes.
That is all anyone canask, and from that noone can make an
?in?ite' diagonal construct at all in practicality.Every
number is on the list because the most they can differ by is
(0..9)/oo.Herc> It is easy to come up with a list of
rationals that have a rational> diagonal. The set I give
above is one such set.> Russell> - 2 many 2 count>
===
>
I give a constructive method of creating a RM computable
number, x,> that is not in S. How can my method fail to
produce x?> Because every x would be of the form (0)11..1,
and because of the> de?ition of S, any such RM number is
contained in S.The de?ition of S leads to contradiction.> S
can not contain every natural number.> You assume that
every string that corresponds to a ?ite number is> in S.
This is not true if my proof is correct.> That was how S was
de?ed : it contains all the strings of the form> (0)11..1
(this can be considered as all unary representations of>
natural numbers).I show that there exists an RM computable
number of the form (0)11...1> that is not in set S. I can
even show how this number can be> constructed by a Turing
machine.RM's are a subset of TMs.> Any RM can be emulated by
a universal Turing machine (UTM).> We are only concerned with
the subset of RM's that output an initial,> ?ite and
contiguous string of 1's followed by an in?ite string of>
0's. The UTM can generate the output tape for each of these
RM's.> These tapes are then read by a second TM I will call a
Comparator> (CTM). The CTM compares each input tape with the
CTM's output tape.> If the input tape has a longer initial
string of 1's, the CTM rewinds> additional 1.After all of the
RM tapes have beed read by the CTM we examine the> output of
the CTM. This tape must contain the representation of a>
natural number.Every tape read by the CTM represents a
natural number.> The CTM output tape contains the
representation of the successor> of some member of S. The
successor of a natural number is a> natural number.S can not
contain a representation of every natural number.> I can
only think of two reasons why I would not be able to
compute> x. > 1) S contains a member with in?itely many
1's> 2) S contains a member with so many 1's that 
adding>
one more 1 results in an in?itely long string of 1's.> If
you can think of another reason why x can not be computed,>
please post it.> 3) for any s in S,
s-with-a-1-added-to-the-back is also contained in> S Not
true. I show how to construct such a number that is not in
S.> Can you show that x is impossible to compute?> As
there can not exist an element which is both member of a set
and> not contained in the set, yes.The de?ition of S leads
to contradiction.> S can not exist.OK, consider the set of
natural numbers, N.Also consider the operation + 1.Is it not
true that for each n in N, n + 1 is also in N ?Now, consider
the function f : N -> RM-numbers :f(0) = (0)f(1) = (0)1f(2) =
(0)11...and so on, with f(n) corresponding to its own
RM-number for each natural number n.Clearly, not all the
different RM-numbers are reached by this function f. Consider
the set F = f(N) of numbers which are. For each element s of
F, we can ?d a unique natural number n such that s = f(n).
This means we can give the inverse function of f, call this
g.Now, for each s in F, there can not be an x which is s
followed by an extra 1, which is not already in F : this
would mean that g(x)=g(s)+1, with g(s) a natural number,
would not be a natural number.If you still consider this line
of thinking to be wrong, please give me the exact x for which
value it goes wrong.-- PentoDe wereld was soep, en het denken
meestal een vork,tot smakelijk eten leidde dat zelden. - H.
Mulisch
===
 OK, consider the set of natural numbers, N.> Also
consider the operation + 1. Is it not true that for each n in
N, n + 1 is also in N ?Maybe.I might not be the person you
want to ask this question to.My proof shows that no set can
contain every computablenatural number. If N does not contain
every computablenatural number, how can we say N contains all
natural numbers?> Now, consider the function f : N ->
RM-numbers : f(0) = (0)> f(1) = (0)1> f(2) = (0)11> ...> and
so on, with f(n) corresponding to its own RM-number for each
natural> number n. Clearly, not all the different RM-numbers
are reached by this function f.> Consider the set F = f(N) of
numbers which are. For each element s of F,we> can ?d a
unique natural number n such that s = f(n). This means we
can> give the inverse function of f, call this g.> Now, for
each s in F, there can not be an x which is s followed by
anextra> 1, which is not already in F : this would mean that
g(x)=g(s)+1, with g(s)> a natural number, would not be a
natural number.Since g() is the inverse of f(), and x was not
generated by f(),g(x) may not be de?ed.x is RM computable so
there exists a RM that will output it.This means f() can not
emulate every possible RM.Russell- 2 many 2 count
===
> OK,
consider the set of natural numbers, N.> Also consider the
operation + 1.> Is it not true that for each n in N, n + 1
is also in N ?Maybe.>I might not be the person you want to
ask this question to.>My proof shows that no set can contain
every computable>natural number. If N does not contain every
computable>natural number, how can we say N contains all
natural numbers?(I haven't been following this thread, so
please make allowances if what I say here is either redundant
or irrelevant.)The standard de?itions of computable are
equivalent to computable by some Turing machine. Now clearly,
each particular natural number K can be generated by some TM;
indeed, it's even computable by a ?ite-state machine (with K
states).Furthermore, the in?ite sequence of all natural
numbers together (e.g., in base 1, 010110111011110111110...)
can be generated by a (non-halting) TM -- the algorithm
repeatedly copies the last string of 1's (which can be done
?itely by temporarily replacing the most recent 1 copied
with some other symbol, to keep track of how much has already
So in what sense isn't every natural number
computable?[snip]-- ---------------------------| B B aa rrr b
|| BBB a a r bbb | | B B a a r b b | | BBB aa a r bbb |
-----------------------------
===
----------------------------- <^> <(?)> <^>
----------------------------- > OK, consider the set of
natural numbers, N.> Also consider the operation + 1.> Is
it not true that for each n in N, n + 1 is also in N ?>
>Maybe.>I might not be the person you want to ask this
question to.>My proof shows that no set can contain every
computable>natural number. If N does not contain every
computable>natural number, how can we say N contains all
natural numbers? (I haven't been following this thread, so
please make allowances if what> I say here is either
redundant or irrelevant.) The standard de?itions of
computable are equivalent to computable> by some Turing
machine. Now clearly, each particular natural number K> can
be generated by some TM; indeed, it's even computable by a>
?ite-state machine (with K states). Furthermore, the in?ite
sequence of all natural numbers together> (e.g., in base 1,
010110111011110111110...) can be generated by a>
(non-halting) TM -- the algorithm repeatedly copies the last
string of> 1's (which can be done ?itely by temporarily
replacing the most recent> 1 copied with some other symbol,
to keep track of how much has already So in what sense isn't
every natural number computable?>Check out this 3 state TM
that counts from 0 upwards in binary. Everysecond position on
the tape represents the number.Hercknock knock knock on the
door, feet still bleeding
===
>message> OK, consider the
set of natural numbers, N.> Also consider the operation +
1.> Is it not true that for each n in N, n + 1 is also in
N ?>Maybe.>I might not be the person you want to ask this
question to.>My proof shows that no set can contain every
computable>natural number. If N does not contain every
computable>natural number, how can we say N contains all
natural numbers? (I haven't been following this thread, so
please make allowances if what> I say here is either
redundant or irrelevant.) The standard de?itions of
computable are equivalent to computable> by some Turing
machine. Now clearly, each particular natural number K> can
be generated by some TM; indeed, it's even computable by a>
?ite-state machine (with K states). Furthermore, the in?ite
sequence of all natural numbers together> (e.g., in base 1,
010110111011110111110...) can be generated by a>
(non-halting) TM -- the algorithm repeatedly copies the last
string of> 1's (which can be done ?itely by temporarily
replacing the most recent> 1 copied with some other symbol,
to keep track of how much has already So in what sense isn't
every natural number computable?Consider all TM's that write
an initial, ?ite and contiguous strings of1'sand then halt.
Let S be the set of all output tapes from these TM's.De?e
another TM I call a comparator (CTM).CTM compares the input
tape to its output tape.If the input tape has more 1's, the
CTM rewindsto the beginning of its output tape, copies the
input tape,We let CTM read all the tapes in S.What is on the
output tape of CTS?The output of CTM must have a ?ite number
of 1's.Every tape read by CTM was ?ite in length.The output
of CTM has exactly one more 1 thansome input tape in S.S can
not contain every possible string thatrepresents a natural
number.Russell- 2 many 2 count
===
comp.theory and sci.math
need to come off the list of newsgroupsfor this thread. Those
groups surely have more important thingsto discuss. : :
Consider all TM's that write an initial, ?ite and contiguous
strings of : 1's : and then halt.OK. There is one of these TMs
for every natnum. : Let S be the set of all output tapes from
these TM's.That's basically the set of all natural 
numbers. :
De?e another TM I call a comparator (CTM). : CTM compares the
input tape to its output tape.CTM *DOES* *NOT* *HAVE* an
output tape.CTM is a TM. In general, TMs DO NOT HAVE output
tapes.TMs have ONE tape. They use it for input.They also
write on it, but that does not make it an output tape.It is
at best an input/output tape. : If the input tape has more
1's,This is ridiculous. The input tape IS the output tape, if
CTMis a TM. It can NEVER have a DIFFERENT number of 1's from
itself. : the CTM rewinds to the beginning of its output
tape, copies the input tape,That is not how TMs work. What
actually happens is that the TM readsat the end, and then
halts. : We let CTM read all the tapes in S.No, we don't,
because S has an in?ite number of tapes, so CTM(if it really
is a TM, which, as you have de?ed it, IT ISN'T,because it has
an output tape) will never ?ish reading allthese tapes. :
What is on the output tape of CTS?That's a typo; CTS does not
exist; do you mean CTM? : The output of CTM must have a ?ite
number of 1's.Right. : Every tape read by CTM was ?ite in
length.Right. : The output of CTM has exactly one more 1 than
: some input tape in S.Wrong.S has in?itely many input tapes
and there is no upper bound onhow long they are. : S can not
contain every possible string that : represents a natural
number.As you de?ed it, it does indeed contain exactly
that;it represents the natnums in unary.-- --- It's dif?ult
... you need to be united to have any strength, but internal
issues have to be addressed. --- E. Ray Lewis, on liberalism
in America <-sOdnVAq4PkNeXGiRVn-sA@comcast.com>
<Xns945E7EF166C5Brobbygoetschalckxstu@134.58.127.12>
<8aGdnb0k8ZEW73CiRVn-jA@comcast.com>
<Xns945E983979A02robbygoetschalckxstu@134.58.127.12>
<q8CdnfPGJ8E4knOiRVn-tw@comcast.com>
<Xns945F37E9E9C54robbygoetschalckxstu@134.58.127.12>
<Bu-dnYP-tIz1JHOiRVn-sA@comcast.com>
<Xns945F820BCBD74robbygoetschalckxstu@134.58.127.12>
<19idnd3Ywuot6HKiRVn-sA@comcast.com>
<bsnvo0$q5k$1@lust.ihug.co.nz>
<V7GdnTOguOLYF3KiRVn-vw@comcast.com>
===
> [ Obviously, every
natural number N is computable by an N-state FSM. ]> So in
what sense isn't every natural number computable? Consider
all TM's that write an initial, ?ite and contiguous strings
of> 1's and then halt. Let S be the set of all output tapes
from these TM's. S is an in?ite set which is also
countable.> De?e another TM I call a comparator (CTM).> CTM
compares the input tape to its output tape.> If the input
tape has more 1's, the CTM rewinds> to the beginning of its
output tape, copies the input tape, We let CTM read all the
tapes in S. What do you mean read all the tapes in S? S has
in?itelymany members. If CTM tries to read all the members
of S, itwill never ?ish.> What is on the output tape of CTS?
Nothing -- CTM never halts, as it never ?ishes reading
itsinput tapes.> The output of CTM must have a ?ite number
of 1's.> Every tape read by CTM was ?ite in length.> The
output of CTM has exactly one more 1 than> some input tape in
S. Non sequitur.> S can not contain every possible string
that> represents a natural number. Non sequitur. Surely
you've had this explained to you in the past,about how some
in?ities are bigger than others, and whichones, and how?
Check Google Groups if you haven't, or don'tremember 
-- I'm
sure there's plenty of tutorial informationon there. Check
sci.math's archives, too.-Arthur
===
What are the parameteric
coordinates of closed loops generated by intersection point F
between b and c and intersection point I between a and c
extensions in a quadrilateral of sides a,b,c,d ( side a
smallest,side d largest,angle th variable/parameter between a
and d )? These occur in a fourbar linkage mechanism . The book
4 Bar Atlas(by Hrones and Nelson)does not include parameteric
equations. TIA.
===
I need to know how to calculate x^y where
y is not an integer. Anyhelp in this matter would be
appreciated.
===
> I need to know how to calculate x^y where y
is not an integer. Any> help in this matter would be
appreciated.If y is rational, i.e. y = a/b, then b th root
(x^a). For example,x^2/3 = cube root of (x^2). Otherwise,
check out:http://mathworld.wolfram.com/Power.htmlLurch
===
> I
need to know how to calculate x^y where y is not an integer.
Any> help in this matter would be appreciated.If y is
rational, i.e. y = a/b, then b th root (x^a). For
example,x^2/3 = cube root of (x^2). Otherwise, check
out:http://mathworld.wolfram.com/Power.html??? I don't see a
mathematical de?ition of x^y on thatpage, nor anything else
that might be what he's lookingfor when he asks how to
calculate it...>Lurch>************************David C.
Ullrich
===
>I have another more general question, if you
don't mind : is there an>intuitive meaning for a nilpotent
group ? Nilpotent groups are closely connected to nilpotent
lie algebras,>which is where the name actually comes from, as
I understand it, and>where nilpotent actually makes some
sense.>As to intuitive meaning, that will depend a lot on
you, I guess.[...]>Others think of them as groups constructed
through central extensions,>which is not a bad way to think
about them either.Some of us just instinctively think,
Product of p-groups. RealisticallyI just think about
p-groups, period, and then mutter something aboutdirect
products at the end of the story. (And if you want an
intuitivemeaning of p-groups, think about layers of small
vector spaces overthe ?ld of characteristic p, forming the
quotients of chains ofnormal subgroups. I don't necessarily
think only of central extensions,by the way; in fact one of
the prototypes I keep in mind is a semidirectproduct of a Z/p
acting on a p-dimensional vector space over GF(p).I'll bet
most people don't think about the dihedral group of order
8this way...)Of course, some of us carry along a related
intuitive notion thatgroup means ?ite group. You'll have to
modify the previous paragraph if you think
otherwise.daveX-Received: (from approve@localhost) by
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id hBSDc1T02534;
===
 [.snip.]J.
Willekens
===
> Notice that the in Noble Verse 57:25 above,
Allah Almighty says clearly We> sent down iron.... and He
didn't say We created iron from earth....> Allah 
Almighty's
claim was very accurate and precise. We sent down> iron.....
clearly states that iron was created outside the earth and
was> brought down by the Will of Allah Almighty for a
purpose, and that is> (material for) Mighty war, as well as
many bene?s for mankind, that Allah> Noble Quran,
57:25)>Hypothetically, if this is proven incorrect, and the
iron did not come down,would you be willing to give up on the
Quran?Bill <vuop8jtg11tge3@corp.supernews.com>
<bsia8e$pbm$1@neptun.beotel.net>
<vuraj3sm7q1e87@corp.supernews.com> X-Cise:
tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate:
admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose:
george_cox@btinternet.comX-Punge: Micro$oft
===
 at 02:12 PM,
Mark K. Bilbo <noem@il.huh> said:>There is no down in space.
There is if you send a goose up. I don't know if 
that's
enough downfor a jacket.-- Shmuel (Seymour J.) Metz, SysProg
and JOATnot reply to spamtrap@library.lspace.org
===
speak
thusly:> at 02:12 PM, Mark K. Bilbo <noem@il.huh> said:>
>There is no down in space. There is if you send a goose up.
I don't know if that's enough down> for a jacket.The 
poor
goose...-- Mark K. Bilbo <vuop8jtg11tge3@corp.supernews.com>
<bsia8e$pbm$1@neptun.beotel.net>
<vuraj3sm7q1e87@corp.supernews.com>
<bskjme$vpe$1@neptun.beotel.net>X-Cise:
tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate:
admin@interspeed.co.nzX-Ecrate: tanandtanlawyers.comX-Pose:
george_cox@btinternet.comX-Punge: Micro$oft
===
 at 07:30 PM,
Goran Jakupovic <jg01495p@kiklop.etf.bg.ac.yu> said:>Like
somebody else already said not just iron, but all atoms
starting>with helium and heavier now in solar system were
products of>supernova.No. Helium is predominantly primordial,
Beryllium and Lithium arealmost entirely primordial. The
conditions for light elementnucleosynthesis are very
different from the conditions for heavierelements. Stars
produce Helium, but they destroy Lithium andBeryllium.
There's more Helium in the Universe now than right afterthe
big bang, but not a lot more.-- Shmuel (Seymour J.) Metz,
SysProg and JOATnot reply to spamtrap@library.lspace.org
===
>
> down is relative like meaning closer to core like when
you get down> from> bed. but the quran has things being
created on earth but for some odd> reason> it diverges with
iron. it says we sent down iron. iron was not created> on>
earth according to quran. how did the quran know that iron
came down?> down> meaning from elsewhere. it was not made
on this earth. god caused iron> to> descend. coincidence?>
> It's untrue. Iron was here when the earth was formed. It
did not come> later.> And if anything, it came up from the
core from volcanoes and the like.you are mistaken in a small
way. iron not even from this solar system. it> came from
other stars and had to land on earth. maybe it was primitive>
molten earth but earth was here. if earth was not here what
iron land on?> simple logics no offense. the miracle is that
quran say many things created> on earth and if this book from
man who forging gods word man just say iron> like people and
animals and plants and mountains was created on earth. but>
for some reason quran treats iron different and say that it
is sent down to> earth. why not just say it was created on
earth? because that would be> wrong. --> saab siddiqui al
mujahed> but you have to change the (a) to @ for it to
workIron is an element, copper is an element, carbon is an
element, and so on.What scienti? basis do you have for your
belief that iron came latter thanthe others? What made up the
earth that the iron landed on?Bill
===
thusly:> thusly:>
right. i did not say it came to earth after it ?ished with
animals> plants> and all that. i only noted that the quran
has many things created on> earth> but for some reason does
not have iron created on earth.> What about all the *other
elements that weren't formed here?what about them? are you
trying to raise an argument ex silentio? if it> turns out on
usenet you never once say your mothers maiden name does that>
necessarily mean you did not know your mothers maiden name? as
for the> elements other than iron i will be agnostic on what
the author of the quran> knew about them for now.So this is
like one of those christian arguments. God failed to
mentionanything *useful like penicillin but seems to have
mumbled something vagueabout iron.-- Mark K. Bilbo
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 at 07:13 PM,
raynand@netzero.net (Jefferson Rourke) said:>Atheism is
simply a lack of belief in gods.No, that's Agnosticism.
Atheism is the belief in the lack of a god.-- Shmuel (Seymour
J.) Metz, SysProg and JOATnot reply to
spamtrap@library.lspace.org
===
> at 07:13 PM,
raynand@netzero.net (Jefferson Rourke) said:>Atheism is
simply a lack of belief in gods.No, that's Agnosticism.
Atheism is the belief in the lack of a god.How about
un-theist. Does that work for you? I think the god-conceptis
a nonexistant mind created reality that it is used as a tool
todupe the gullible.I think religious mysticism is a form of
mental illness and madnessthat is unique to the human mind on
this planet. This still work foryou?I think the god-concept
and religion have held back the development ofthe human race
by 2000 to 3000 years. Instead of looking for thenonexistent
afterlife the human race should be working to build abetter
life here and now. If not for religion and the god-concept
itis a possible that we could be on other planets by now and
havelifespans of over 500 years. The loss and waste of
potential in thehuman race because of the god-concept is
immense.Still working for you?If all of this does not ?
within your de?ition parameters pleaselet me know about my
ill-de?ed thought processess.Jefferson RourkeLaissez-Faire!
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 at 08:02 PM,
raynand@netzero.net (Jefferson Rourke) said:>How about
un-theist. Does that work for you?It's not a question of
whether it works for me; it's a question ofwhether it works
for the English language. The English terms areagnostic and
atheist. What would work for me is honesty, which you arenot
exhibiting.>I think religious mysticism is a form of mental
illness and madness>that is unique to the human mind on this
planet. Then you were lying when you stated that Atheism is
simply a lack ofbelief in gods. You have gone beyond not
believing in a god tobelieving that there is no god. >I think
the god-concept and religion have held back the development>of
the human race by 2000 to 3000 years.Perhaps it did. And
perhaps it had the oposite effect. Do you haveevidence, or is
it just a matter of faith for you?>Still working for you?Nope.
But if your faith works for you, . . .>If all of this does not
? within your de?ition parameters please>let me know about
my ill-de?ed thought processess.You might start with post
hoc ergo propter hoc.-- Shmuel (Seymour J.) Metz, SysProg and
JOATnot reply to spamtrap@library.lspace.org
===
speak thusly:>
at 08:02 PM, raynand@netzero.net (Jefferson Rourke) said:>
>How about un-theist. Does that work for you?It's not a
question of whether it works for me; it's a question of>
whether it works for the English language. The English terms
are> agnostic and atheist. What would work for me is honesty,
which you are> not exhibiting.Nope. Wrong.Some of the
colloquial connotations that have accreted to the words
aresimilar to what you're claiming but colloquial meaning
shifts around allthe time.Atheism is lacking belief in any
gods. The word was coined to mean that.That is what atheists
use the word to mean. That is the meaning.-- Mark K.
Bilbo
===
thusly:> at 07:13 PM, raynand@netzero.net
(Jefferson Rourke) said:>Atheism is simply a lack of
belief in gods.> No, that's Agnosticism. Atheism is the
belief in the lack of a god.How about un-theist. Does that
work for you? I think the god-concept> is a nonexistant mind
created reality that it is used as a tool to> dupe the
gullible.Actually, there's no point to trying to change
terms. Theists would trashany term used.And atheism actually
*is un- or maybe more accurately non- theism.Since a- means
without. As in amoral means *without morals ascontrasted to
immoral which is *not moral. The in- pre? meaningopposite of
or not.What they try to claim is atheism is, is more something
you might callintheism (to coin a word). That would be
opposite of theism. Atheism works ?e. Fits the way we do
things in the language (hence theword was brought into the
language to mean without theism). Theistmisunderstanding or
even deliberate obfuscation notwithstanding...-- Mark K.
Bilbo
===
> thusly:> at 07:13 PM, raynand@netzero.net
(Jefferson Rourke) said:>Atheism is simply a lack of
belief in gods.> No, that's Agnosticism. Atheism is the
belief in the lack of a god.How about un-theist. Does that
work for you? I think the god-concept> is a nonexistant mind
created reality that it is used as a tool to> dupe the
gullible.Actually, there's no point to trying to change
terms. Theists would trash> any term used.And atheism
actually *is un- or maybe more accurately non- theism.> Since
a- means without. As in amoral means *without morals as>
contrasted to immoral which is *not moral. The in- pre?
meaning> opposite of or not.What they try to claim is atheism
is, is more something you might call> intheism (to coin a
word). That would be opposite of theism. Atheism works ?e.
Fits the way we do things in the language (hence the> word
was brought into the language to mean without theism).
Theist> misunderstanding or even deliberate obfuscation
notwithstanding...Hey Mark:I was just trying to have a bit of
fun with the guy and see what hisresponse would be. I thought
I had the de?ition right the ?st timeand I was writing off
of the top of my head without double checking.Hope you had a
happy Winter Solstice,Jefferson Rourke
===
thusly:> thusly:>
> at 07:13 PM, raynand@netzero.net (Jefferson Rourke)
said:>Atheism is simply a lack of belief in gods.>
> No, that's Agnosticism. Atheism is the belief in the lack
of a god.> How about un-theist. Does that work for you? I
think the god-concept> is a nonexistant mind created reality
that it is used as a tool to> dupe the gullible.>
Actually, there's no point to trying to change terms. Theists
would trash> any term used.> And atheism actually *is un-
or maybe more accurately non- theism.> Since a- means
without. As in amoral means *without morals as> contrasted
to immoral which is *not moral. The in- pre? meaning>
opposite of or not.> What they try to claim is atheism
is, is more something you might call> intheism (to coin a
word). That would be opposite of theism. > Atheism works
?e. Fits the way we do things in the language (hence the>
word was brought into the language to mean without theism).
Theist> misunderstanding or even deliberate obfuscation
notwithstanding...Hey Mark:I was just trying to have a bit of
fun with the guy and see what his> response would be. I
thought I had the de?ition right the ?st time> and I was
writing off of the top of my head without double checking.No
big thing. I was just rattling along on an interesting (to me
at least)subject.I can't help it. I'm moving more and 
more
into linguistics as (if thingshold together) will be my next
?ld.Words fascinate me to no end...-- Mark K. Bilbo
===
speak
thusly:> at 07:13 PM, raynand@netzero.net (Jefferson Rourke)
said:>Atheism is simply a lack of belief in gods.No,
that's Agnosticism. Atheism is the belief in the lack of a
god.No, atheism is lacking belief in gods.We know, we're
atheists.-- Mark K. Bilbo
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 at 11:42 AM,
Saab Siddiqui <p@O.g> said:>im snipping mr metz points that i
have no response to at this>time.Good; that is proper quoting
style for Usenet. You'll see peopleis generally something to
avoid and not to imitate.>not that star collide with earth
but that matter from star like iron>collide with earth.The
Earth was formed by the accretion of matter. Much of that
matterwas Iron. There was already a substantial amount of
Iron when theEarth was very small. The material that fell
later included a lot ofelements besides Iron, and had no
higher percentage of Iron than theinitial material. Given the
outgassing of lighter elements, theprimordial proto-Earth
probably had a higher concentration of Ironthan the new
material falling on it did.-- Shmuel (Seymour J.) Metz,
SysProg and JOATnot reply to spamtrap@library.lspace.org
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 at 04:56 PM,
Mark K. Bilbo <noem@il.huh> said:>The process is that iron
forms in stars. The stars go nova and eject>material which
includes iron. No. Iron forms in signi?ant quantities only
in stars that gosupernova. A simple nova is not hot enough or
dense enough.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot
reply to spamtrap@library.lspace.org
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>Not just
large, but dense. A white dwarf star is about the size of>the
Earth,Well, I prefer a yellow dwarf, but would rather that it
remain a safe93,000,000 miles away ;-)>I do wonder what would
happen if the impacting object was a neutron star,Lethal. The
details would depend on the size, but I imagine that
theradiation would kill us before we had a chance to observe
the rest.-- Shmuel (Seymour J.) Metz, SysProg and JOATnot
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 at 01:51 AM,
Steve Knight <wooooly@sonic.net> said:> Great. Some camel
ing, rag head, sand muncher, At least he is not a racist
xenophobe like you. *PLONK*-- Shmuel (Seymour J.) Metz,
SysProg and JOATnot reply to
spamtrap@library.lspace.org
===
>How do you prove what the
(imaginary) zeroes of the Fibonacci