mm-461
===
Russell Easterly says...>The set of all natural numbers is
recursive by fiat.>Why bother to include computability in the
definition?>A human must decide if x is a member of this
set>before x can be given to the TM.Right. The notion of
recursive set is only nontrivialfor proper subsets of the
naturals. For the complete set,it is recursive by definition.
===
Subject:
: Quadratics and transformationIt seems to me that I
should be able to reasonably easily get a generalidea of the
type(s) of transformations that have been applied to x^2just
by looking at the finished quadratic. That is, if I have 3x^2
+ 7x + 2, can I easily relate the coefficientsand constant to
the dilations, reflections and translations that thisfunction
represents compared to x^2 ? I have tried thinking about it as
a combined function, that is,f(x)=x^2, g[f(x)]=3x^2 + 7x + 2
and trying to find g(x). Well, I haven'ttried with this
example but one I tried earlier turned into a horriblemess and
didn't give me what I wanted - I assumed that g(x) is simplythe
inverse of g[f(x)] so maybe that's why it got so messy ? It
also seemed to be a lot of work!I'm aware that I could
determine the turning point and work out thetranslations and
that the roots would indicate what horizontal dilationhas been
applied. I expect that if I calculate two other points I
wouldalso be able to determine reflection and vertical
dilation. Of course Icould always graph it but I'm hoping to
achieve what I want frominspection of the description of the
function (is that the rightterminology?). Any and all help
appreciated.Ivan.===
===
Subject:
: re:JSH: Why am I so important?I
see what he's up to, we'll fill this thread with posts and
he'll goto somewhere else and say that he's important and will
have thisthread to back him up, and the chain goes on forever.
Don't reply tohim. ===
===

Subject:
: Re: HELL :-) (Was:
Re: the anticlassicalist }{ ii: the spectre continues)>
*MATHEMATICS IS FULL OF TRICKS*French even allows you the
conjugation :ma th.8ematiqueta th.8ematiquesa
th.8ematiquenotre th.8fme .88 ticsvotre th.8fme attiqueleurre,
t'es mat, hic !> *REDUCING TOPICS TO THE SCIENCE OF NUMBER IS
NUMEROLOGY*I'd call that accounting.===
===
Subject:
: Re: pentiamond
rep-tiles, continued> That leaves these two:> ____ __/> /____
/___/> I see no reason that these should not be rep-tilable.
Neither do I see> a rep-tiling of either! Does anybody?I now
see a reason that the pentiamond on the right should not
berep-tilable. Suppose you have such a rep-tiling. Look at the
littlepentiamonds along the edge of the big one, starting at
the top vertexand following the edges on the right and bottom.
You will find thatall the boundaries between the little
pentiamonds have the same orientationrelative to the edge of
the big one: _______________You cannot escape from this
pattern, even when you pass a 120-degreeangle. When you reach
the lower left vertex, you're stuck.No solution yet on the I-,
or straight, pentiamond. It can't be doneon a scale of 9x9 or
smaller.web: <http://www.monmouth.com/~colonel/>===
===
Subject:
:Re: errors in an argument>Let's just look at the probability
of a protien made up of>100 amino acids forming by chance. All
the amino acids>have to line up just right, and the probability
of this>happening in any given try is 1/100! or 10^(-158).>
There are only 20 different amino acids.> The more subtle
error is that nobody says that life required> exactly this
protein to be produced by a random permutation. > There are
probably a huge number of different proteins that> could do
more or less whatever this one does. And its production> would
not be a matter of random permutations, but rather of> the
gradual accumulation of small improvements. You might> look up
genetic algorithms: this is not just a theoretical> construct
but a method which is actually used to obtain a >
nearly-optimal solution to a difficult problem by simulating >
the process of evolution.> That said, I think it's fair to add:
my impression, as an> outsider in this field, is that while
evolution itself (for > organisms that already have the usual
genetic machinery) has pretty> good mathematical models, we
are still far from understanding> at a quantitative level how
that genetic machinery might have > arisen in the first
place.Robert, of course, mentioned exactly the errors I was
thinking of,with the second being that we don't know how many
possibleproteins there are with 100 amino acids in them. To
say that foreach one, the probability is essentially 0, and so
the collectiveprobability for the set of proteins is 0 is quite
a mistake.If I were trying to explain the error to someone, I
would probablyuse the analogy of rolling a 20 sided dice 100
times and recordingthe result. By the proof above, one can't
do this, since theprobability of any particular sequence is
(essentially) 0.I guess what I would really like to know is
how poorly I shouldthink of the person who made the argument.
Here are somequalifiers: He boasts of his degree in
mathematics. Also, in histalk, he indicated that he would try
to teach us how to thinkcritically about what scientists say.
On the other hand, he isnow in his mid 60's, so he is far
removed from his schooling.Finally, almost certainly (this is
a guess of mine) he did notformulate the proof himself, but it
had been presented to him.He liked the argument and missed the
flaws.Should I not be too critical of his wishful thinking
that thisproof actually means something?By the way, I
certainly agree completely with Robert'slast
paragraph.John===
===
Subject:
: Re: Big Rip NY Science TimesIs it my
imagination, or are this idiot's ramblings becomingeven *more*
delusional and illucid than before?===
===
Subject:
: Re: How to
diagonalize a Hermitian matrix>Hello group, Could somebody
please tell me a numerical algorithm to>diagonalize a
Hermitian matrix H, so that I end up not only with
the>eigenvalues perched along the diagonal, but also with the
unitary>matrix U that conjugates with H to diagonalize it. I
should probably say what I really want. I have
a>positive-definite Hermitian matrix H, and I need its matrix
square>root. The above is simply the approach that occurred to
me for>obtaining that square-root.> There are available
algorithms to do this diagonalization.> Any good library
should have such.> I am assuming that by square root you mean
the Hermitian> square root. If you just want a matrix A with
H=A*A~, the> conjugate transpose of A, the Cholesky
decomposition is the> fast way to do it.> One can use Newton's
method to obtain the square root of a> matrix; the rate of
convergence is not bad. There are other> ways as well. I
suggest you ask the question in> sci.math.num-analysis.think
that is what I am looking for. I will try on the group
yousuggest.Achava===
===
Subject:
: Re: Sets That Resemble
Derivatives Somewhatthere's a general theme that some people
have explored that says thatthe fact that certain sorts of
differential operators obeyleibniz-like identities is
conceptually subordinate to the fact thatcertain sorts of
boundary operators in geometry and topology obeyleibniz-like
identities, roughly (or sometimes precisely, depending onthe
particular concept of boundary being used) boundary(x X y)
=(boundary(x) X y) + (x X boundary(y)).-- [e-mail address
jdolan@math.ucr.edu]===
===
Subject:
: Big
Riphttp://qedcorp.com/destiny/under constructionDARK MATTERS
SURROUND DARK ENERGYwith my comments, additions and physics
correctionsTwo big stories from the world of physics may
portend the arrival of new weapons of mass destruction far
more powerful and compact than atomic bombs. In recent years
it has been discovered that our universe is being blown apart
by a mysterious anti-gravity effect called dark energy.
Mainstream physicists are scrambling to explain this
mysterious acceleration in the expansion of the universe. Some
physicists even believe that the expansion will lead to The Big
Rip when all of the matter in the universe is torn asunder -
from clusters of galaxies in appears to be made of two
unknowns - roughly 23% is dark matter, an invisible source of
gravity, and roughly 73% is dark energy, an invisible
anti-gravity force. Ordinary matter constitutes perhaps 4
percent of the universe. Recently the British science news
journal New Scientist revealed that the American military is
pursuing new types of exotic bombs - including a new class of
isomeric gamma ray weapons.That was an original idea of mine
in 1963 at Cornell and I discussed it with Hans Bethe. That is
one of the reasons Ron Bullough invited me to Harwell in 1966.
No doubt others thought of it but probably later. I thought of
it while at Tech/Ops in Lexington, Mass working for George
Parrant Jr.Unlike conventional atomic and hydrogen bombs, the
new weapons would trigger the release energy by absorbing
radiation, and respond by remitting a far more powerful
radiation. In this new category of gamma-ray weapons, a
nuclear isomer absorbs x-rays and remits higher frequency
gamma rays. The emitted gamma radiation has been reported to
release 60 times the energy of the x-rays that trigger the
effect.Gamma-ray weapons could trigger next arms race 19:00 13
August 03Exclusive from New Scientist Print Edition. Subscribe
and get 4 free issues.An exotic kind of nuclear explosive
being developed by the US Department of Defense could blur the
critical distinction between conventional and nuclear weapons.
The work has also raised fears that weapons based on this
technology could trigger the next arms race. The explosive
works by stimulating the release of energy from the nuclei of
certain elements but does not involve nuclear fission or
fusion. The energy, emitted as gamma radiation, is thousands
of times greater than that from conventional chemical
explosives.The technology has already been included in the
Department of Defense's Militarily Critical Technologies List,
which says: Such extraordinary energy density has the potential
to revolutionise all aspects of warfare. Scientists have known
for many years that the nuclei of some elements, such as
hafnium, can exist in a highnergy state, or nuclear isomer,
that slowly decays to a lownergy state by emitting gamma rays.
For example, hafnium-178m2, the excited, isomeric form of
hafnium-178, has a half-life of 31 years. The possibility that
this process could be explosive was discovered when Carl
Collins and colleagues at the University of Texas at Dallas
demonstrated that they could artificially trigger the decay of
the hafnium isomer by bombarding it with lownergy X-rays (New
Scientist print edition, 3 July 1999). The experiment released
60 times as much energy as was put in, and in theory a much
greater energy release could be
achieved.http://www.newscientist.com/news/news.jsp?id=
ns99994049I was thinking in 1963 of a gamma ray laser pumping
a nuclear isomeric transition. Bethe at the time said it
wouldn't work and basically discouraged me working on
it.Bekkum continued:In the summer of 2000 I contacted Nick
Cook, the former aviation editor and aerospace consultant to
Jane's Defence Weekly, the international military affairs
journal. Cook had been investigating black budget super-secret
research into exotic physics for advanced propulsion
technologies.Uh Oh :)I had been monitoring electronic
discussions between various American and Russian scientists
theorizing about rectifying the quantum vacuum for advanced
space drive. Several groups of scientists, partitioned into
various research organizations, were exploring what NASA calls
Breakthrough Propulsion Physics - exotic technologies for
advanced space travel to traverse the vast distances between
stars. Partly inspired by the pulp science fiction stories of
their youth, and partly by recent reports of multiple radar
tracking tapes of unidentified objects performing impossible
maneuvers in the sky, these scientists were on a quest to
uncover the most likely new physics for star travel. The NASA
program was run by Marc Millis, under the Advanced Space
Transportation Program Office (ASTP). Joe Firmage, a Silicon
Valley entrepreneur, who at age 28 had found risen to CEO of a
three billion dollar internet firm, began to fund research in
parallel with NASA. He hired a NASA Ames nano-technology
scientist, Creon Levit, to run the International Space
Sciences Organization,Joe did that because I suggested it. I
introduced Creon to Joe.Cook was intrigued by the apparent
connections between various private investors, defense
contractors, NASA, INSCOM (American military intelligence),
and the CIA. While researching exotic propulsion technologies
Cook had heard rumors of a new kind of weapon, a sub-quantum
atomic bomb, being whispered about in the dark halls of
defense research.I think that must have come from me regarding
J. P. Vigier's tight atomic states with experiments in Beograd
by Z. Maric and G. Dragic. But how did Cook hear about that?
We brought Vigier to ISSO in San Francisco several times along
with physicist Gennady Shipov from Moscow. That story with
photographs of Vigier and the group is in my autobiography
Destiny Matrix. Dragic A, Maric Z, Vigier JP; Phys. Lett. A
265 (2000) 163. New quantum mechanical tight bound states and
'cold fusion'. Creon Levit and Vigier met with Maric in
Budapest in 2000.Bekkum who is one of my on line students
continued:Sub-quantum physics is a controversial
re-interpretation of quantum theory, based on so-called pilot
wave theories, where an information that the predictions of
ordinary quantum mechanics could be recast into a pilot wave
information theory. Recently Anthony Valentini of the
Perimeter Institute has suggested that ordinary quantum theory
may be a special case of pilot wave theories, leaving open the
possibility of new and exotic non-quantum technologies. Even
thought rumors of a sub-quantum bomb may be purely fantasy
It's not fantasy. It might not work, Maric and Dragic in
Beograd, while not ostensibly trying to make a weapon by any
means, were trying to test Vigier's basic theory of the
spatially extended electron, which I think is basically a
correct idea and fits my own ideas including why the electron
appears to shrink to less than 10-16 cm under high resolution
imaging (i.e. scattering) and how the electric charge
distribution is contained by the strongly attractive zero
point energy exotic vacuum dark matter core
(Abraham-Becker-Lorentz-Poincare stress problem of 100 years
ago). This only works in the Bohm hidden or extra variable
interpretation. That is, a classical spatially extended
electric charge distribution is unstable. It explodes under
its own self-repulsion. This is why physicists had to
postulate a point electron because they did not understand
that the strong gravity attraction of the positive zero point
pressure in a possible state of exotic vacuum would hold the
charge together. As Herbert Frohlich told me at UCSD in La
Jolla in 1966 the basic thing wrong with physics is the idea
of the point electron. The bad idea of the point electron
gives the infinite energy in quantum electrodynamics. Richard
Feynman told me in his office at Cal Tech in 1968 that
infinite renormalization is a shell game, and it is a scandal
in physics that no one could do better than what he had done.
They did not know 100 years ago that 1/3 or so of the universe
was this kind of exotic vacuum. For example, there is a huge
sphere of exotic vacuum of w = -1 positive pressure that holds
our galaxy together preventing our solar system from going off
into space on its own. This sphere looks like w = 0 cold dark
matter from our vantage point. What works on this large scale
also works on the small scale of the single electron (and all
the charged lepto-quarks). A neutrino has some mass and is
simply a micro-geon of pure zero point energy with positive
pressure. there is no question that physicists seriously
contemplate a phase transition in the quantum vacuum as a real
possibility. The quantum vacuum defies common sense, because
empty space in quantum field theory appear and disappear far
too quickly to be detected directly, but their existence has
been confirmed by experiments that demonstrate their influence
on ordinary matter.A major component of the physical quantum
vacuum consists of virtual electrons frothing and bubbling at
the Fermi surface edge of the Dirac negative energy sea. This
is because of the Pauli exclusion principle that only none or
one electron per quantum state. A virtual electron pops out of
the vacuums Fermi surface leaving a hole behind. The hole is
the virtual positron. The result is a virtual
electron-positron pair. However, the virtual electron and the
virtual positron attract because they have opposite charges
and they are exchanging virtual photons. Therefore, some of
them form a more stable bound state. An enormous number of
these virtual pairs Boseinstein condense into the same center
of mass quantum wave packet forming the Vacuum Coherence Field
(AKA Inflation Field). This is a dynamic steady state of
detailed balance in which there is a continual inflow and
outflow of virtual pairs into and out of this giant quantum or
macro-quantum superfluid. Essentially this is a vacuum phase
transition, similar to the BCS transition from a normal metal
to an electrical superconductor, from the globally flat
micro-quantum electrodynamic vacuum without any gravity at all
to the curved macro-quantum electrodynamic vacuum with emergent
gravity. Einsteins field equation of general relativity can be
derived from the phase wiggles and ripples in the robust
stable macroscopically occupied center of mass quantum wave
packet of the bound state of the virtual electron-positron
pair. The exotic vacuum dark energy and dark matter are simply
the amplitude wiggles and ripples of this same virtual pair
quantum wave packet. The wave packet spreads over the entire
3D space of the post-inflationary bubble on which our
Hubble-horizoned universe is located along with an infinity of
parallel American. If the world hologram idea is correct, take
the surface area of the expanding Hubble sphere that is the
causal retarded boundary of 3D space of our past light cone at
Earth and divide it by the quantum of area. That gives us the
number of Bekenstein-Shannon c-bits and explains the arrow of
time (AKA Second Law of Thermodynamics) of increasing
thermodynamic entropy in terms of the dynamical expansion of
the 3D space of the universe. Lenny Susskind calls this
DeSitter Space.Such research should be forbidden!Too late.
Pandoras Box is open. Schrodingers Cat has jumped out of it.In
the early 1970's Soviet physicists were concerned that the
vacuum of our universe was in fact only one possible state of
empty space. The fundamental state of empty space is called
the true vacuum. Our universe was considered to reside in a
false vacuum, protected from the true vacuum by the wall of
our world. A change from one vacuum state to another is known
as a phase transition. This is analogous to the transition
between frozen and liquid water. Lev Okun, a Russian physicist
and historian recalls Andrei Sakharov, the father of the Soviet
hydrogen bomb, expressing his concern about research into the
phase transitions of the vacuum. If the wall between the
vacuum states was to be breached, calculations showed that an
unstoppable expanding bubble would continue to grow until it
destroyed our entire universe! Sakharov declared Such research
should be forbidden! since there was always the possibility
that an experiment might accidentally trigger a vacuum phase
transition.British Astronomer Royal, Sir Martin Rees, Master
of Trinity College, and Director of the Cambridge University
Institute of Theoretical Astronomy on Madingley Road where
Stephen Hawking works discusses all this in Chapter 9 of his
important book Our Final Hour.Could the wall of our universe
be breached from within? The amount of energy required to
punch a hole through the wall appeared to be enormous, and no
known natural physical phenomena, even the most energetic, had
punched through either. A recent report commissioned to examine
potential dangers at the Large Hadron Collider, one of the next
the best of our existing knowledge. Others are not so certain,
however. At least one of the Russian physicists I had
corresponded with was said to have been a former associate of
Andrei Sakharov. He strongly hinted at new theories the
Russians had developed which allow for the manipulation of the
fundamental constants of nature, but he never revealed more
than a sketch of his ideas. He claimed that a breakthrough was
within reach, perhaps within five years Who was that? Not
George Ryazanov?Recent theoretical explorations may suggest
another approach to the physics of the vacuum. The invisible
gravitating dark matter could be the other side of the
invisible dark energy coin, and that suggests the possibility
of manipulating the vacuum for energy release.Now this is my
original idea that you got from our communications over the
past few years. I am the only physicist in the world today, as
far as I know who has suggested this and has already published
it in my two books of 2002 so its in the official record at
the Library of Congress.If a controllable parameter could be
found to mediate the balance between the invisible dark
forces, the result would unleash the vacuum energy of creation
in all of its awful power and majesty. If it were possible to
control the dark sides of the force then spacetime, the arena
where everything we know takes place, could be bent and
twisted with infinitely greater ease than was ever suspected.
This would open Pandora's box to everything from vacuum energy
weapons of mass destruction (capable of destroying the
universe!) to spacetime warp drives and time machines.Exactly,
the above is the thesis of all my books since 2002 at least.A
quick survey of the international electronic archive of
physics papers at www.arXiv.org shows that research into the
vacuum of spacetime for energy production is alive and well.I
do not think that is true. You need to cite specifics here.
There are lots of flakey new age papers on free energy on the
Web written by people without any real credentials but they
are not on www.arXiv.org which is not even allowing competent
fringe papers in controversial topics like cold fusion. So
what exactly are you thinking of here? Indeed, Paul Ginsparg,
who controls the archive, does not even allow Carlos Castro to
publish conservative competent papers on Clifford Algebras
which are not fringe at all!Most authors are independent
researchers struggling with limited funding and resources, yet
their theoretical results suggest that somewhere in Nick Cook's
black world, a major breakthrough has already taken place. Most
likely the United States and Russia are in the lead, but China,
France, Ukraine, Iran, India and Saudi Arabia all have
scientists actively pursuing the fundamental physics that
determine the fabric of our reality, and are seeking the
theory and the means to access the enormous energies locked
inside of the vacuum since the creation of the universe. Even
if the black budget world has yet to unleash the enormous
potential of vacuum energy, there are signs that those in
power may have begun to take notice. Dr. Harold Puthoff, a
scientist with strong government connections, who has
previously worked on classified projects for the CIA, is a
major proponent of vacuum energy physics. Nick Cook's book,
The Hunt for Zero Point, and his recent stories on zero point
energy in Jane's Defence Weekly have also brought attention to
the dangers and military potential of vacuum research. The
American intelligence community financed so-called psychic
spies for over twenty years and through four presidential
administrations. It is highly unlikely that they would ignore
the potential of the quantum vacuum. Dr. George Chapline, of
the Lawrence Livermore National Laboratory, and Dr. Jack
Sarfatti in San Francisco, knew each other in the sixties in
La Jolla, have independently proposed that the quantum vacuum
may unstable to the formation of coherent virtual processes.
Sarfatti suggests that gravity is an emergent property
determined by the physics of the vacuum. His idea is to find a
means of directly interacting with the physics of the vacuum
that controls the shape of spacetime. Such a possibility would
be consistent with the reported success of EvgenyPodkletnov,
the Russian scientist who is experimenting with spinning
superconducting disks. Podkletnov's most recent papers report
the appearance of a mysterious coherent beam of gravity like
radiationwith a measured force of 1000 G. In an interview on
BBC radio, Nick Cook pointed out one immediate application of
the Podkletnov beam - the destruction of missiles and
satellites in flight or in orbit around the earth. Cook showed
the BBC internal documents from Boeing, the American aerospace
contractor, proving their interest in Podkletnov's
research.This beam stuff I am suspicious of. Of course if the
experiment is good, I have to think more about it. I am not so
sure if Podkletnovs experiment is any good and has been
replicated.The connections between Podkletnov's results, and
the kind of vacuum research explored by Sarfatti, beginning in
1999 at the International Space Sciences Organization are the
latest threads in a trail that most likely originates in cold
war disinformation, a game played by East and West against
each other. Glasnost has shifted the balance of partnerships
and the positions of the players, but not the stakes of an
outcome that would leave the world with even more prolific and
powerful weapons of mass destruction.That is true, as shown in
Destiny Matrix, however you leave out the most important
evidence -UFOs!whole business are the connections. Although
Nick Cook never revealed the identity of his deep throat
contact called Dr. Dan Marckus in the book The Hunt for Zero
Point, there was no question that the Podkletnov results had
played a major part in fitting together the pieces of the
puzzle. The amount of interest was in Podkletnov's reports by
NASA, Boeing, and others in the international arena of
aerospace and military research communities was evidence that
there was more here to explore than the latest musings of the
intellectual elite. The truth is that a fundamental theory of
gravity at the scales of subatomic nuclear physics does not
exist. The fact is that no one understands the nature of the
gravitational field at very small scales. In fact gravity has
barely been probed much below one millimeter. Every attempt to
unify the physical theories of gravity with the well-known
standard model physics of electromagnetism, and the strong and
weak nuclear forces, has failed. More importantly there has
been recent progress in the exotic areas of mainstream
research, such as superstring theory, which suggest new kinds
of physics, which might support explanations for Podkletnov's
impulse gravity effect. One of the current fads in theoretical
physics involves large extra dimensions of space that allow a
much stronger version of gravity to leak off the membrane
world of our ordinary three dimensions. The large dimensional
picture allows for the well known forces of electromagnetism,
and the strong and weak nuclear forces, to be confined to a
three dimensional brane-world floating in a higher dimensional
space. Gravitons, the are able to slip off of our brane-world,
which explains why the gravitational force is so much weaker
than the other forces that hold matter together. Gravitons
could be exchanged between our brane-world and another brane
floating nearby in the same higher dimensions.The Sarfatti
picture offers a more direct interaction with the new physics
than the brane world ideas. Sarfatti's vision is to find a
means of using electromagnetic fields in the Josephson effect
to couple to the virtual electron-positron pair giant coherent
condensate inflation field inside the vacuum that controls the
shape of spacetime to the real electron pair giant coherent
condensate of a control high temperature superconductor. UCBs
Ray Chiao has a similar idea using a superconductor to
transduce electromagnetic far field waves to gravity waves
with high efficiency conversion. Sarfatti wants to do the same
thing with non-propagating electromagnetic and gravity near
fields. One wonders if the black budget world may have already
produced some of the technology needed to explore and test
these new realms.Not a chance. They are clueless about the
theory. They are still stuck in Hal's PV model and Bernie
Haisch's zero point ideas, which will never, in my opinion
fly. They are not asking the right questions and do not have
the right idea in their minds. It is my belief, until I see
evidence to the contrary, that I am the only physicist on the
planet today who is doing real theoretical work directly
relevant to the achievement of practical metric engineering
anchored in the now observed reality of dark energy. All my
work is public. I would love to be proved wrong on this
especially by Hal Puthoff, but I am not holding my breath.;-)
Extraordinary claims require extraordinary proof. Everything
else I have seen is either on the wrong track asking the wrong
questions like the work of Puthoff, Haisch, Ibison & Rueda for
example, which at least is real physics that has proved itself
wrong in Ibison's PV cosmology paper, or else the claims are
patently obvious nonsense that Feynman called Cargo Cult
Science. There is also the Russian torsion work of Akimov and
Shipov and I am not prepared to make a definitive statement on
that, as the issue is not simple because of several factors
some political. One must be careful there to separate Shipov's
theoretical work from claims made about practical devices
including weapons applications. I do, however, agree with you
that there is a real issue here as defined in Ch. 9 of Martin
Rees's Our Final Hour.Hal Puthoff coined the term metric
engineering for flying saucer technology. Hal has been working
on this problem for many decades and has held high USG security
clearances and has been privy to reliable information that the
saucers are real and are alien. Otherwise he would not be
working on the problem. However, Hal's theories, both of the
zero point energy and of the gravity field will not solve the
problem because they are too naively based and do not ask the
right questions. The basic physics required for this task is
way beyond the depth of Puthoffs self-described engineering
approach and can be found in Rovelli's new book on quantum
gravity.===
===
Subject:
: Re: No Set Contains Every Computable
NaturalRussell Easterly says...> The usual notion of
recursive set is that a set S is> recursive as a *subset* of
the naturals if there is a> Turing machine T such that for
any *natural* number n> (in unary notation, to be
specific)>The set of all natural numbers is not a proper
subset of>the set of all natural numbers.But it *is* a subset.
A recursive subset.> T(n) halts and outputs 1 <-> n is an
element of S> T(n) halts and outputs 0 <-> n is not an
element of S> If x is some input that is not a representation
of a natural> number, then there is no constraint on what
T(n) does.>According to this definition, the set of all
natural numbers>is recursive because there exists a TM that
doesn't even>read the input and always outputs a 1.Right.> The
set of all natural numbers is recursive by fiat.No, not by
fiat. It is a trivial consequence of a general definition--the
definition that Daryl McCullough has reproduced for you.> Why
bother to include computability in the definition?There is no
separate definition for the case of all natural numbers.There
is one definition that applies to all subsets of the set of
naturalnumbers. The set of all natural numbers (henceforth N)
is a subset ofN. We apply the defintition to N and find out
trivially that it isrecursive.It is not a trivial definition.
It is quite profound. But it istrivial that N is recursive.> A
human must decide if x is a member of this set> before x can be
given to the TM.You are seeking to create a new definition and
theory of computability.Good. You want to consider a larger
context of inputs other than just natural numbers. For
example, you want to include infinite stings as acceptable
inputs. Fine. I would agree with you thatif we enlarged the
universe of acceptable inputs from just N to a largerset
containing infinite strings, then N would probably not be
recursive in any reasonable new definition of the word. You're
right,a TM would not be able to conclude that an infinite
string of 1's isnot a natural number. I would say that in this
enlarged context, thatN would be recursively enumerable but not
recursive.But your problem is that you are using terms with
agreed upon definitionsby the mathematical community (like
recursive) incorrectly. You haveyour own personal definition
that you have failed to state. This is whymathematicians are
arguing with you and pointing out your obvious errors. If you
would like to re-define recursive by all means do so.But make
your new definition known. It would also be good to giveit
another name to avoid confusion, like Easterly-Recursive.Until
you do this, you will probably continue to make false
statementsthat are easily refuted.Remember, mathematics is
precise.By the way there are already in existence more
generalized recursion theories,which deal with recursion on
other domains besides N.-Leonard Blackburn> Russell> - 2 many
2 count===
===
Subject:
: Re: Quadratics and transformation> It seems
to me that I should be able to reasonably easily get a general>
idea of the type(s) of transformations that have been applied
to x^2> just by looking at the finished quadratic.> That is,
if I have 3x^2 + 7x + 2, can I easily relate the coefficients>
and constant to the dilations, reflections and translations
that this> function represents compared to x^2 ?Yes. 3(x^2 +
7/3*x) + 2, complete the square inside the parentheses. Soyou
end up with a(x-b)^2 + c. b is translation along the x-axis, c
alongthe y-axis, a is the dilation factor. If a is negative,
that's areflection.===
===
Subject:
: Re: Understanding taylor
expansion for sineJo Christensen> I've been trying to
understand the taylor exapnsion for the sine> function,>
sin(x) = x - x^3/3! + x^5/5! - x^7/7! + ...> I've been told
that it uses the fact that e^(i*x) = cos(x) + i*sin(x),> and I
can easily work out the taylor series for e^x, but I'm not
sure> how to use these two pieces to find the expansion for
sin(x).> Any help would be appreciated.> Jo
Christensen---Outgoing mail is certified Virus Free.Checked by
AVG anti-virus system (http://www.grisoft.com).===
===
Subject:
: Re:
e is transcendental (was: classes of transcendental numbers ?
by support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id i1IHZOR13942;>Why do I even try?
Are you even TRYING to learn? Or are you >so closed-minded
that you can't be bothered?><snip>So? The real part of
exp(i pi) is cos(pi), and its imaginary part>is sin(pi), so
all you are saying is that cos(pi) = -1 and >sin(pi) = 0, and
we were already aware of these facts. There is >NO reason to
conclude that exp(i pi) = 0. > I,have a reason , with my due
respect.> Panagiotis Stefanides>Yes, but you DON'T tell us what
your reason is. You can't expect us to >accept your claims
without giving support for those claims. So what >possible
reason could you have for expecting us to agree with your
claim >that exp(i pi) = 0?>The reason is simply that
exp(ipi0=-1 should be accompanied>by the statement that this
is the real part solution.>Is it fair?>No, that is not a fair
comment. exp(i pi), the complex number, is>equal to -1, the
complex number. There is no need to appeal to the >real part.
Also, what does exp(i pi) = -1 is the real part solution>mean?
You look like you are using terminology in a manner not
>recognized in mathematics.>-------------->e^[i*pi] ,as
accepted ,is a phasor. >In most of the relevant fields of
mathematics, e^[i pi] is a complex >number.>It is only fair
to state that its >polar representation is :>e^[i*pi] = MOD 1
, ARG 180 .>That is Arg 180 degrees, not just Arg 180. And so
what? That does >not lead to your assertion that exp[i pi] =
0, a result for which >you have given absolutely no support.
Why don't you just give up?> At the moment, they (the Time
Lords) are far from being all-powerful.> That's why it's been
left up to me and me and me.> quote by: Patrick Troughton in
The Three Doctors>------->I, have made myself very
explicit.>My original question of the implication>of the
imaginary component: e^[ipi]=j*0 (to the related
proof)>Complex Notation, chapter 12 ,Electrical Technology 3RD
ED.>Edward Hughes Longmans ,page 338:>Stares:>
OA*=OB+jOC=OA(COStheta+jSINtheta)>* Symbols representing
phasors are printed in bold face italics, while those
representing only magnitudes are printed in ordinary
italics,..>Here is very clear the difference between PHASOR
and MAGNITUDES>I know the difference between a complex number
and its modulus >(or magnitude). You don't have to explain the
difference to>me.>[which(MAGNITUDES) I, referred to as REAL
PART or IMMAGINAR PART,>The magnitude of a complex number is
generally not equal to either>its real or imaginary part. How
could you claim that it is?I, gave a reference:Complex
Notation, chapter 12 ,Electrical Technology 3RD ED.Edward
Hughes Longmans ,page 338:States:
OA*=OB+jOC=OA(COStheta+jSINtheta)* Symbols representing
phasors are printed in bold face italics, while those
representing only magnitudes are printed in ordinary
italics,.. What are these magnitudes then ?> I should have
stated COMPONENTS ].>Perhaps you mean that (the imaginary
part of e^[i pi]) = 0, in which>case you are correct, but you
have had a lot of problem expressing >yourself, especially in
view of the way that you initially made>the claim by stating
that e^[i pi] = 0, which you described as the >imaginary part
solution, using a terminology that nobody but you >knows.>
e[i*theta]=COStheta+iSINtheta>I know that it is true that
exp[i theta] = cos(theta)+i sin(theta). >It follows that exp[i
pi] = -1. Incidentally, e[i theta] = i e theta, >unlike what
you have written.>thetas could be given and calculations could
be performed>for numerical evaluations.>And for other results
as well.>In books is stated that it is FORMULA>and also terms
such as evaluate:(-1+i*sqrt[3])^10>Are these not solutions to
problems?>No. Evaluate (-1+i sqrt(3))^10 is a problem for
which you can>get a solution using exp(i theta) = cos(theta)+i
sin(theta). This>does not mean that exp(i theta) = cos(theta)+i
sin(theta) is itself>a solution. You need a problem before you
can describe anything as>a solution.Of course ,this crops up
when theta is substituted by a given angle.>Nobody knows what
you mean when you make a statement like exp[i pi] = -1>is the
real part solution of exp[i pi] = -1, or that exp[i pi] =
0>is the imaginary part solution of exp[i pi] = -1. I asked
you to >explain your terminology but you haven't bothered.>I,
doubt it but ,still I, exlpained it anyway.>I did ask you. And
you did not explain your bizarre terminology. There>was no
problem, hence there is no solution, whether real part
or>imaginary part.How else would do You call them,I, said:I
should have stated COMPONENTS>As I, exlained earlier, but I,
am not the kind of not >being bothered>This was not a quote
from me. Why are you pretending that it was?You are correct,
this was my quote.>When I take the imaginary part of the
equation exp[i pi] = -1, I get>sin(pi) = 0, a fact which is
already known to us.>Have you thought also of the fact that
if you have exp[i pi] = -1 (your >real part solution) and
exp[i pi] = 0 (your imaginary part solution),>then you could
conclude that 0 = exp[i pi] = -1, and get a
contradiction?>Well tthere is always the possibility of an
answer such as >MULTIVALUED , examplum gratias:>It is known
that the exponential function is single-valued (in fact,>it is
known that the exponential function is entire).> e^[2i*pi]=1 >
ln[1]=0=2i*pi>The logarithm has a branch point at 0. The
exponential function does>not have a branch point. The analogy
is invalid. Because the >exponential function is single-valued,
my objection here still stands.>I, make use of the comlex
notation , but still ,I, have my >natural questions ,and do
not accept everything for granted.>But you shouldn't start
making up mathematics to suit yourself. This is not my
intention,believe me.>I, give an example in the form of
question:>It is required that COS[-i]+i*COS[-i] be
evaluated>so that is its Modulus and Argument>be evaluated
(NUMERICALLY].>cos(-i) = cosh(1), so cos(-i)+i cos(-i) =
sqrt(2) cosh(1) exp(i p/4).>The magnitude is sqrt(2) cosh(1),
which is approximately 2.18225,>and its argument is pi/4,
which is approximately 0.7854. >Alternatively, the argument is
exactly 45 degrees.I, Panagiotis
Stefanides>--------------===
===
Subject:
: Re: help with solutions
for three questions from the past contest> 3. A boat with an
ill passenger is 7.5 mi north of a straight> coastline which
runs east and west. A hospital on the coast is 60> miles from
the point on shore south of the boat. If the boat starts>
toward shore at 15 mph at the same time an ambulance leaves
the> hospital at 60 mph and meets the ambulance, what is the
total distance> (to the nearest 0.5 mile) traveled by the
boatand the ambulance? ans:> 62.5> Boat .> |> | > 7.5| z> |
> Shore |___________________________________.Hospital> x 60-x>
Hours along distance z at 15 mph must equal hours along
distance> 60-x at 60 mph.> And at constant speeds, distance =
speed * time.I'll be curious if that's the best solution. (It
may be the one expected.)A better solution would be to
minimize the time to the hospital. That's astandard calculus
question (usually phrased in terms of a swimming pool),with
the additional proviso that if the ambulance wouldn't be to
the meetingpoint at the time the boat gets there, then this
solution should be used.Of course, I wouldn't be a proper
kibbitzer if I actually worked out thedetails before posting
this criticism.Dare we point out that, if it's a true
emergency, the coast guard sends ahelicopter?===
===
Subject:
: Re:
How to choose a matrix PNo D!=Eso P can't be an identity
matrix>Now I am thinking of this question.>Suppose >Y1=D -
P'EP >Y2=D >where D,E,P : nxn matrix>||P||=1>How to choose P,
so that >rank Y1 >= rank Y2 >and >||Y1||>= ||Y2|| > The
identity matrix would seem to work...> Maybe you have some
more requirements you're not telling us.===
===
Subject:
: Teaching
philosophyI need to write a description of my teaching
philosophy. In order todo so accurately, I think I need to
write at greater length and ingreater detail on this topic
than any hiring committee will want to read.Moreover, I need
to present my views accurately but, somehow, in sucha manner
as not to vitiate the consideration of my application.That
being the case, can someone please tell me what my
teachingphilosophy is?Ignorantly,Allan
Adlerara@zurich.ai.mit.edu************************************
***************************************** ** Disclaimer: I am
a guest and *not* a member of the MIT Artificial **
Intelligence Lab. My actions and comments do not reflect ** in
any way on MIT. Moreover, I am nowhere near the Boston **
metropolitan area. **
**************************************************************
***************===
===
Subject:
: Re: proving integrability of
binomial numbersEn el
mensaje:5bbb5f5d.0402172032.54e70ed3@posting.google.com,Achava
Nakhash, the Loving Snake <achava@hotmail.com>
escribi.97:Ignacio Larrosa Ca.96estro
<ilarrosaQUITARMAYUSCULAS@mundo-r.com> PS - I have never
heard of Pascal's triangle referred to as> Tartaglia's
triangle. I only know of Tartaglia as a contributor to>
solving the general cubic. Is this because I mostly have
access to> American and other English language sources?Google
[Tartaglia triangle] or [Tartaglia triangulo], or of
course,[Tartaglia triangolo] ...At secondary school, I always
saw Tarataglia's triangle (well, actually'Tri.87ngulo de
Tartaglia').-- ===
===
Subject:
: Re: question about periodic
function by support1.mathforum.org (8.11.6/8.11.6/The Math
Forum, $Revision: 1.9 primary) id i1IHh7M14607;>If you have
two periodic functions f(t) and g(t) with periods a and
b,>then the function f(t) + g(t) is periodic with period c
where c is the>least-common-multiple of a and b. Here the
first period is 1/.018=500/9 nd>the second period is 1/.02=50.
So the least-common-multiple is 500, i.e.>1/.002. That should
be the period of the combined function.How to prove the
theorem? Any hint is welcome. TIA.===
===
Subject:
: Re: strain
softening spring>i.e. the equation I am trying to solve is as
follows:>y'' + cy' + [ko*e^(-alpha*t)]y = 0>Any help would be
most appreciated.>sincerely>Paul
Joseph>(pjoseph@excite.com)Maple gives a solution in terms of
Bessel functions. I posted a filecalled DE.pdf showing the
solution at:http://math.asu.edu/~kurtz/de/--Lynn===
===
Subject:
:
Re: intercept length when a random line intersect with
ellipsoid shell or parallelepiped shell with fixed thicknessEn
el
mensaje:477a8527.0402180800.756b523e@posting.google.com,ChenHS
<chenhuisu@hotmail.com> escribi.97:> If a random line
intersects with a ellipsoid shell or parellelepiped> shell
with a fixed thickness, how much is the statistical intercept>
length?You must define in a precisse way how the random line is
choosed. The answerdepend on it and it can be very different.--
===
===
Subject:
: Re: CAN ANYONE HELP ME?????CAN ANYONE HELP ME?????>
The 'caps lock' key is on the left-hand side of the keyboard,
above> 'shift' and below 'tab'. It should look like this:>
+--------+> | |<--- | <-- tab key> | --->| |> +--------+-+> |
Caps | <-- caps lock key> | Lock |> +--------+-+> | / |> | ||
| <-- shift key> +--------+> Over on the right-hand side of
the keyboard, above the numeric keypad,> should be three LEDs,
labelled (from left to right) 'Num Lock', 'Caps> Lock' and
'Scroll Lock'> Press your 'caps lock' key until the LED marked
'Caps Lock' is no longer> lit.> Now you are ready to post on
usenet.1.DETERMINE DOMAIN OF THE FUNCTIONf(X)=X3/X2+2X+1> What
happens when x^2 + 2x + 1 = 0?Excellent! I (along with Baghdad
Bob and most readers) assumed he meant(x^3/x^2) + 2x + 1
(following the standard rules for order of operations),but you
ferreted out the probable real meaning, which for OP's
clarificationshould have been written x^3/(x^2 + 2x + 1). FIND
EXTREME FUNCTIONS f AND DRAW ITS GRAPHextreme points? extreme
points of the function?2.BY NEWTONS METHOD SOLVE
EQUATION:X3-X-2=0Your book or your notes have Newton's method
(or Newton-Raphson iteration).Google also gets about 37,000
hits (only 4,780 for Newton-Raphson), but someof these are
advanced treatments. You can still get good info, but you
haveto look for elementary treatments.===
===
Subject:
: Re: the
anticlassicalist }{ iv: from maps to logic> -=-=-=-=-= from
maps to logic =-=-=-=-=-=-> Now let us look at interpretting
this arrow symbol, such as in process,> categorial inclusion,
rule refinement, or progressions of state. We want> to be able
to evaluate logical relationships between objects. So we look
to> define on a recognition map new objects defined by any two
objects which> allows us to collectively compare two objects
logically. The first one of> these, which can be called the
least upper bound or disjunction of two> states 'a' and 'b',
written (a / b) which obeys the rules that> a -> (a / b)> b ->
(a / b)> forall x in the lattice with a -> x and b -> x, we
have (a / b) -> x> where it is important to differentiate when
these are derivationally found> on a lattice (objects already
defined through previous axioms with the> appropriate
universal property derivable) and when the existence of new>
objects is being defined (for each new definition requires new
consistency /> completeness / etc. checks).You mean it provides
a natural framework for conjectures ?> dualityDoesn't tell much
about that. Duality is negation in boolean algebras, isn't it
?===
===
Subject:
: Re: question about periodic function>If you have
two periodic functions f(t) and g(t) with periods a and
b,>then the function f(t) + g(t) is periodic with period c
where c is the>least-common-multiple of a and b. Here the
first period is 1/.018=500/9 nd>the second period is 1/.02=50.
So the least-common-multiple is 500, i.e.>1/.002. That should
be the period of the combined function.>How to prove the
theorem? Any hint is welcome. TIA.If f(t) has period a then
f(t + na) = f(t) for any integer n, right?Same idea for g: g(t
+ mb) = g(t) for any integer m.Hmmmm....how could I find an
integer that works for both.....?--Lynn===
===
Subject:
: Re:
Teaching philosophy> I need to write a description of my
teaching philosophy. In order to> do so accurately, I think I
need to write at greater length and in> greater detail on this
topic than any hiring committee will want to read.> Moreover, I
need to present my views accurately but, somehow, in such> a
manner as not to vitiate the consideration of my application.>
That being the case, can someone please tell me what my
teaching> philosophy is?What kind of philosophy? Metaphysics?
Epistemology? Ethics? Aesthetics? Kantian? Pragmatic?
Platonic? Aristotelean?Bob Kolker===
===
Subject:
: Re: Teaching
philosophy>I need to write a description of my teaching
philosophy. In order to>do so accurately, I think I need to
write at greater length and in>greater detail on this topic
than any hiring committee will want to read.>Moreover, I need
to present my views accurately but, somehow, in such>a manner
as not to vitiate the consideration of my application.>That
being the case, can someone please tell me what my
teaching>philosophy is?>Ignorantly,>Allan
Adler>ara@zurich.ai.mit.eduI would be happy to do so, but I
only know how to take multiple choicetests. What are the
choices?:-)--Lynn===
===
Subject:
: Re: How many different
resistances with n resistors?Originator:
tchow@lagrange.mit.edu.mit.edu (Timothy
Chow)>tchow@lsa.umich.edu schrieb:> The Puzzle Corner in MIT's
magazine Technology Review raised the> question of the number
of distinct resistances achievable by connecting> ten unit
resistors.>Tony oletti (seq. A048211) considers only cases
that can be reduced by>applying the formula for
parallel/series. If you have 6 resistors there is
1>configuration where this is not possible. I suppose we get
an additional>resistance value.What's the 6-resistor
configuration you have in mind?And has anyone calculated the
answer to the problem if we don't assumeseries/parallel
circuits?-- Tim Chow tchow-at-alum-dot-mit-dotduThe range of
our projectiles--ven ... the artillery---however great,
willnever exceed four of those miles of which as many thousand
separate us fromthe center of the earth. ---Galileo, Dialogues
Concerning Two New Sciences===
===
Subject:
: Re: errors in an
argument by support1.mathforum.org (8.11.6/8.11.6/The Math
Forum, $Revision: 1.9 primary) id i1IIGEO17931;>Should I not
be too critical of his wishful thinking that this>proof
actually means something?His argument is an overly
sophisticated version of one of the most basic arguments
against evolution, one that I can't rule out. How does super
evolved, complex, fine tuned to the millimeter biological
structures appear when their evolutionary benefit seems to be
absolutely nothing even with 90% or 95% of them installed.
People used to point out that eyes are one of these mechanisms
from the very beginning,but that, I think, is a bad example
because any ability to sense this way, even on a veryprimitive
level, must be a huge advantage.My favorite is birds : How can
one evolve into flying ? So many things have to happen inorder
to create functional wings it seems impossible any awkward
animal developing inthis direction would survive it's cost, or
any failing flying attempts. The evolutionistarguments sound
totally absurd : they ran fast or jumped from tree to tree and
got anadvantage slowly developing wings. Sounds like absolute
voodoo to me. ===
===
Subject:
: Re: the anticlassicalist }{ v:
universal truths> And on these lattices we can make more
explicit our theories of negation.> On sci.logic, mitch
recently gave a very interesting post on Stone algebras.> When
we want to model uses of various definitions, we need to show
them as> an axiom set. Stone algebras study the axiom set:>
~(a) / ~(~(a)) <--> TBtw, can complex numbers model it ? Let's
see~ : r exp(i s) --> r/2 exp(i s/2)a = 1~(a) = -1/2~(~(a)) =
i/4let's say T is 0...Well, then I am sure a Lorentz-Moebius
transform can save the situation, I mean, provide a natural
interpretation to the same formula graphics. The Moebius group
is strictly 3-transitive, and z / w <---> ucan pass as denoting
-the- moebius transform that swaps z and w while keeping u
invariant. Now idea if the thing can mimick a lattice any
further, though.Isn't the symmetric group attached to the
boolean algebra in such a manner that extremely transitive
(but non-symmetric) groups shine as interesting analogues
?===
===
Subject:
: Re: . 53ab2750 is an anonymous user .
charset=iso-8859-1X-Mimeole: Produced By Microsoft MimeOLE
V6.00.2800.1165AHahahahaha....AHAHAhahahaha...ahahaha.........
you too get three attaboys for your reply !Life's a bowl of
cherries, pits and nuts and
all....ahahahahaha.......ahahahahanson> Yeah. My niece used to
laugh like you ... when she was 2 years oldehhhhh,
hmmmmmm.........ahahahaha......Your retort has a pH 12+ for
some reason. Is it due to an after effect from ~ ....after
chemo, I, Peter Brown was never the same?Cheer up, Peter,
*laugh*, that's all that's there, that's all you got
left!Life's a bowl of cherries, pits and nuts & all....Do you
git it? git it git it?hanson, bonvivant and commentator on all
walks of
life...AHAhhahaha..Ahahahahahaha..........
AHAHAHHAHHAAHAHAHHAHa........ahahahahaha..===
===
Subject:
: Re:
Teaching philosophy>I need to write a description of my
teaching philosophy. In order to>do so accurately, I think I
need to write at greater length and in>greater detail on this
topic than any hiring committee will want to read.>Moreover, I
need to present my views accurately but, somehow, in such>a
manner as not to vitiate the consideration of my
application.>That being the case, can someone please tell me
what my teaching>philosophy is?>Ignorantly,>Allan AdlerA more
serious answer:I promise to cheerfully accept students who
don't have theprerequisites and can't add fractions without a
calculator or solvelinear equations. I will have them work in
groups so, hopefully,someone in each group can do the work. I
will be their best friend sothey don't think of me as a
teacher. We will have nice pleasantclasses where we all
discover things together so nobody feels left outor has their
selfsteem bruised. Before each exam I will give out apractice
exam ****very**** similar to the real exam so they will
knowwhat to expect. I will be very generous with part credit
for thosethat still don't get it, even giving re-takes if
necessary. I**guarantee** a very high success rate in my
classes. There will be somany A's and B's that I will get
awesome student evaluations. Studentswill flock to my classes
because I am such a great teacher. The Deanwill notice the
improvement in my department's success rate and willprobably
increase it's funding. Over time, as my methods catch on,even
the University administration and maybe even the Regents
willhold our department up as a shining example of what is
being done toimprove higher education.--Lynn===
===
Subject:
: Re:
HELL :-) (Was: Re: the anticlassicalist }{ ii: the spectre
continues)*MATHEMATICS IS FULL OF TRICKS*> French even allows
you the conjugation :> ma th.8ematique> ta th.8ematique> sa
th.8ematique> notre th.8fme .88 tics> votre th.8fme attique>
leurre, t'es mat, hic !*REDUCING TOPICS TO THE SCIENCE OF
NUMBER IS NUMEROLOGY*> I'd call that
accounting.lol:-)mitch===
===
Subject:
: Re: Simple numbersWell what
is that algorithm?Hallo;Well, p(n) is a n-th prime but if
n=123456 how do I know what is the123456-th prime? I need the
correct number (for exampel 1-st prime is ->2)!> Hallo,> I
know that every simple number (beside 2 and 3) has its
oneformula :>S.N => 6*n +/- 1 for some n , but not any
n={1,2,3,...+} in formula 6*n+/- 1>gives> the simple number
(for example: n=6 => 6*6-1=35 , 35 is not simple,>5,7|35)> By
simple number I think you must mean prime.> so my question
is:> If n={1,2,3,...} what is the formula which gives the
simple numbers>(beside> 2 and 3) for any n?> Evidently you
don't consider p(n) is the n'th prime to be aformula.>Any>
particular reason?>Computational complexity?> Given any n,
there is a simple algorithm that computes p(n). You couldspend
a> lifetime trying to make the algorithm more efficient, but
the basicalgorithm> is quite simple.===
===
Subject:
: Re:
[arccos((sqrt(5)-1)/2 )] / pi is irrational> I will use this
to show that > if cos x = (1 + sqrt(5))/2 then x is
transcendental. (Hence, it is > irrational.)> But that
wasn't the original question. The original question > was
whether x / pi is irrational. A whole nother kettle of fish.
To make this question actually look interesting, note that if
cos(x)=( (1+sqrt(5)) / 4 ), then x/pi is equal to 1/5. (Also,
the original question has sqrt(5)-1 instead of sqrt(5)+1 .
Again, cos(x) = ( (sqrt(5)-1)/4 ) implies x/pi = 2/5 . The
question is to show that this 2/5 value turns transcendental
when that '4' on the denominator changes to a '2.')J> So sorry
for the bleeps in my non-solution.> Yes, the original question
did involve (sqrt(5)-1)/(2 *p)i not> sqrt(5)+1 and Lindemann's
lemma and> my proof are only true if x ne 0. Of course e^0 =
1.> I will go back to the drawing board and see if my original
proof can> be fixed.> Sorry for the whole nother kettle of
fish!> Ray SteinerWell, I finally got the
solution(hopefully).I don't have time to post all the details
right now, but here is abrief outline of the steps:1). Use
cos(2*pi/5)= (-1 + sqrt(5))/4 to find all the roots of unityin
Q(exp(2*pi*i/5)).(There are 10 of them.)2). Now consider
cos(pi*r)= (-1 + sqrt(5))/2.Then, by the usual formulas of
trigonometry, we getcos(2*pi*r)= 2-sqrt(5)andsin(2*pi*r)=
2*sqrt( -2+sqrt(5)).If r were rational, cos(2*pi*r) +
i*sin(2*pi*r) (*)would be a root of unity in
Q(exp(2*pi*i/5)).3). Since (*) doesn't match any of the
answers of step 1, r cannot berational.Hence, r is
irrational.Hope this does the job!Ray Steiner===
===
Subject:
: Re:
Full Beal Conjecture (revised)> Sorry for the mistake. My
proof can be obtained atDid you win the dosh?-- ===
===
Subject:
:
Re: strain softening spring> I am a Civil engineer who has
been working for the last 20 years in> the field as a result
of which my math is somewhat basic. I have a> good handle of
linear systems, but am taking my first baby steps> with
non-lineear systems.> I have read up a few books on non-linear
systems looking for acanned solution for a problem I am trying
to solve. But while I see> a lot of stuff on the Duffing
equation I don't see anything that> resembles the equation I
am trying to solve - a mass-spring-damper> system where the
spring softens exponentially.> i.e. the equation I am trying
to solve is as follows:> y'' + cy' + [ko*e^(-alpha*t)]y = 0>
Any help would be most appreciated.Well, if you want an
intuitive (albeit approximate) solution, you shouldstart by
considering the relative size of the parameters. Specifically,
youhave ordinary damping due to the 'c' parameter, and then you
have theexponential softening, i.e. the 'alpha' parameter.
These both have units ofinverse time and you should start by
determining which range of values youare interested in, e.g. c
<< alpha and/or c > alpha. In either of thesecases, the time
scales are very different, and you can therefore make
someassumptions. For instance, if alpha is very small, then
the softeninghappens on a very slow time scale and therefore
the spring constant isalmost, well, constant. You can
therefore assume that the solution is anordinary damped
oscillation and simply replace k0 in the solution with k0
*e^(-alpha*t).Another line of approach is to try to eliminate
the damping by substitutingy(t) = z(t) * e^(-beta*t) for some
suitable chosen value of beta.A third idea is to make a vector
plot. Introduce z=y' and for each point(y,z) draw a vector
whose direction is (y',z'). Try it first with alpha =
0,because then the vectors are independent of time. You will
see spirals thatlead to the point (0,0). When alpha is
nonzero, the vector plot istime-dependent, but you might be
able to say something qualitative aboutthem anyway, e.g.
whether they spiral faster or slower towards (0,0).===
===
Subject:
:
A newbie's question -- about real numberTwo real numbers r1 =
0.89, r2 = 0.889999999..... (with infinite 9s), ===
===
Subject:
:
Re: A newbie's question -- about real numberlisong@iastate.edu
asks this about real numbers:>Two real numbers r1 = 0.89, r2 =
0.889999999..... (with infinite 9s), Theoretically NO.In real
life measurement or result calculations in which only two
significantfigures are reasonable, then r1=r2. The context
must be appropriate. G C===
===
Subject:
: Definition of Separable
Space (basic topology question)Let A be a subset of (X,T).
Then A is dense in X iff for every nonmptyopen subset U of X,
A / U != {}.I have seen two definitions of 'separable
topological space':a) (X,T) is separable if there exists A
<subset> X, where A is countable anddense in Xb) (X,T) is
separable if there exists A <proper subset> X, where A
iscountable and dense in XGiven def (a), its simple to prove
that all countable spaces X areautomatically separable since
the subset X is countable and non-triviallyintersects every
nonmpty open subset. However, this won't work given def(b). I
first became suspicious when I saw a proof that all countable
spaceswere separable that seemed ridiculously complex in
comparison to theseemingly obvious 1 step proof above. I later
found definitions ofseparable like def (b).l8r, Mike N.
Christoff===
===
Subject:
: Re: A newbie's question -- about real
number Adjunct Assistant Professor at the University of
Montana.>Two real numbers r1 = 0.89, r2 = 0.889999999.....
(with infinite 9s), Both expressions are taken to represent
the same real number. Thefirst is the real number 8/10 +
9/100.The second is taken to represent the limit of the
infinity series8/10 + 8/100 + (9/10^3 + 9/10^4 + 9/10^5 +
....)Since Sum_{i=3}^{infinity} 9/10^i =
9/10^3*(Sum_{i=0}^{infinity} 1/10^i),and the series (1/1 +
1/10 + 1/100 + 1/1000 + ...) converges to1/(1-(1/10)) =
1/(9/10) = 10/9, we have that the original infiniteseries8/10
+ 8/100 + (9/10^3 + 9/10^4 + 9/10^5 + ....)converges to8/10 +
8/100 + 9/10^3*(10/9) = 8/10 + 8/100 + 1/100 = 8/10 +
9/100,which is equal to r1.-- =====Arturo
Magidinmagidin@math.berkeley.edu===
===
Subject:
: Cauchy's
Interlacing TheoremHi folks,I'm looking for a proof of the
following...Let A = [a0 a^h ; a A2] >= 0 be a NxN hermitian
matrixwhere a0 is a scalar > 0 and A2 is the lower-right
N-1xN-1(principal) submatrix of A.According to Cauchy's
interlacing theorem (1829 I guess),the N Eigenvalues gamma_i
of A and the N-1 Eigenvalues alpha_iof A2 interlace, i.e.,
gamma_1 >= alpha_1 >= gamma_2 >= ...Interestingly, having
played a little bit with MATLAB,also the Eigenvalues gamma_i
of A and the Eigenvalues beta_iof the N-1xN-1 matrix
A2-a*a^h/a0 seem to interlace. However,this seems to be true
only for nonnegative-definite A.Does anyone have a proof, or
reference to a proof for that??Initially, I wanted to show
that if A >= 0 with rank A >= 2,then also A2-a*a^h/a0 >= 0;
but, then I discovered Cauchy'stheorem... ;)(If rank A = 1,
then A2-a*a^h/a0 yields the N-1xN-1 zero matrix,and if rank A
>= 2, then rank A2-a*a^h/a0 >= 1, at least thisis easily
seen).Once the above assumption had been proven, A2-a*a^h/a0
>= 0would directly follow as beta_N-1 >= gamma_N >=
0(comparing the smallest Eigenvalues)Christian.===
===
Subject:
: Re:
Collatz Conjecture : Symmetry question.> Every time I post on
this problem I get beat up so please be kind. I hope that
wasn't directed at me. I never beat anyone. It may seemthat
way to Ernst Berg, but I was just trying to help him by
offeringconstructive criticism. Apparently some people can't
get past the criticism part and get all bent out of shape to
the point where theyignore the constructive part. But that's
Ernst's problem.> I have found many properties in the
landscape of this problem and the> one that has most of my
interest the gross amount of non-regularity> tightly coupled
to large a domain of symmetry. I have noticed 2 forms> of
symmetry in this problem which have me thoroughly intrigued,
but do> not know where to go with these notions at this
point.> Symmetry 1;> I have noticed that given any tree
pattern of any size, it is exactly> duplicated infinitely many
times. For example, take the entire tree> starting at A and
extend out all branches to a depth of any N. Then> that
pattern taken generically is repeated at an offset of A +>
n*(constant value) with (n = 1, 2, 3, ... infinity), with the
constant> value is a function of N.That seems reasonable since
any sequence vector evaluates to a simpleformulaa' = (X*a -
Z)/Ywhich means there are always an infinte set of integer
solutions to everysequence vector.> Symmetry 2;> I have also
noticed that all values at N (here I include all> rationales
and integers by taking all possible paths from A to N,>
keeping NON_INTEGER branches in this process) produces N' sets
of a> collective 2^(N-1) number of values where the number of
items in each> set follows the Binomial Coefficients or
Pascal's Triangle (which ever> you normally call it). If you
take each of these N' sets of values and> calculate the
ordered set difference (N'') that the difference sets> (N'')
are constant and equal for all N independent of A.I'm not sure
I follow this, but since there are only two operations (3x+1and
x/2) every number has two possible ancestors, so carried to N
levels(and ignoring whether the ancestors are integers) there
would certainlybe 2^N numbers at distance N.> This means that
using only the N'' sets, all sets at N in groups of> N', thus
all values at N, for any A, can be calculated without>
extending the tree from A to N through all the intermediate
N's.> In other words, all values of absolute stopping distance
of N, for any> A, can be calculated directly with out
traversing the tree to find> each of those values.But there
are vastly more total numbers at a given level than there
areintegers, so although this is true, I'm not sure it leads
anywhere.For example, I recently found all the integers that
are at level 84(from 1):179,441,377If I included all the
potential non-integer ancestors, I would have had tosift
through 2^84 or19,342,813,113,834,066,795,298,816numbers to
find which ones are integers. In this case, it is much quicker
to traverse the tree from level 1 up to level 84. I have a
program that canbuild each level from the previous one. Since
it only generates integerresults, it is much more efficient
than testing each binary permutaionof the two operations
(provided you record the integers you collected from the
previous level).It would be nice to be able to just start at
an arbitrary level and findall the integers. So although it
can be done in theory, it can't be donein practice for high
levels.Unless you have some means of determining which of the
2^84 numbers at level84 are integers without having to iterate
through them all.> Is this interesting to anyone? Yes.> Or
better yet, Is this anything of interest for this problem.I'm
not an expert, so I can't say.===
===
Subject:
: Re: A newbie's
question -- about real number>lisong@iastate.edu asks this
about real numbers:>Two real numbers r1 = 0.89, r2 =
0.889999999..... (with infinite 9s), >Theoretically NO.Hence
the rationals are uncountable! ===
===
Subject:
: Graph Theory
Textbook I'm an undergraduate, and this summer I will be
participating in anREU for discrete math and combinatorics. I
am looking for a goodgraph theory textbook to learn the basics
from. I have seen the booksthat Dover (which of course are very
cheap) has to offer on thesubject, but unfortunately they seem
too elementary. Is the SpringerGTM Graph Theory by Reinhard
Diestel any good? Any suggestions arehighly
welcomed.===
===
Subject:
: Re: errors in an argument> My favorite is
birds : How can one evolve into flying ? So many things> have
to happen in order to create functional wings it seems
impossible any> awkward animal developing in this direction
would survive it's cost, or> any failing flying attempts. The
evolutionist arguments sound totally> absurd : they ran fast
or jumped from tree to tree and got an advantage> slowly
developing wings. Sounds like absolute voodoo to me.That
depends on the size of the animal. If the animal is small
enough,the terminal velocity can result in a non-lethal impact
with theground. (I believe the threshold is around the size of
a mouse or acat.) Incremental levels of control over the
landing point then have anobvious benefit. --
http://members.iquest.net/~panoptes/039 53 36 N / 086 11 55
W===
===
Subject:
: Re: 3 Squares Covering 1 Circle> For b, what does
a C3v (invariant under 120 deg rotation) > symmetric
arrangement yield? (Place the outermost edge of> the squares
tangent to the circle.)Here's a similar situation that does
better than the 80.4%with part (a). Use this 120 deg symmetry,
but place the diagonalsof the square along the radii. The
optimum placement of thesquares (they can be shifted along the
radii) covers 83.0% ofthe circle's area. This is the best that
I've found, but I don'tknow if it's the best possible.> To be
a little more precise, the area is 2.60808705... I have an
exact> form, but it's pretty messy. This happens when the
diagonal of each> square extends .13980931 past the center of
the circle (along the> diameter).Oh, that's a nice
configuration to consider!I checked your result, and I agree
with your area Ac = 2.60808705,although the extension of the
square beyond the center of the circle isa little off: it
should be rc = 0.13977317. (I would guess that youobtained the
value of Ac to single precision, which yields only half
theprecision for rc).For general r (distance diameter extends
beyond the circle's center),the area A of the resulting figure
can be found as follows:Let u = sqrt(2) - r, v = sqrt(2 - u^2),
d = (u - v)/2, andp = (1 + u v)/2, then A = 3 (p - sqrt(p) d -
(3 + sqrt(3)) r^2/2 + Arcsin(d)).This area is maximized when r
equals the critical value rc: rc = (19 - 5 sqrt(3) - sqrt(176 -
86 sqrt(3))) / (26 sqrt(2)).The formula for Ac = A(rc)
simplifies greatly (as one would expect).Let dc = rc (3 +
sqrt(3))/2. Then Ac = 3 (1 - sqrt(2) dc + Arcsin(dc)).To
summarize the results so far, the best result posted for case
(a)(no overlap) is(a) pi/6 - 1/2 + sqrt(3)/4 + sqrt(55)/8 +
Arccos(3/8) = 2.57003584 = 81.8067816% of the circle,and the
best result posted for case (b) (overlap allowed) is(b) Ac
(see above) = 2.60808705 = 83.0179893% of the circle.-Jim
Ferry===
===
Subject:
: Re: A newbie's question -- about real number>
Hence the rationals are uncountable! Nice try, but no cigar.
The set of decimal sequences that terminate in all 0's or all
9's is countable.===
===
Subject:
: Re: A newbie's question -- about
real number> Theoretically NO.> In real life measurement or
result calculations in which only two significant> figures are
reasonable, then r1=r2. The context must be
appropriate.Theoretically YES, and given that we discuss
mathematics here, that is the only meaningful
answer.===
===
Subject:
: Re: Definition of Separable Space (basic
topology question)> Let A be a subset of (X,T). Then A is
dense in X iff for every nonmpty> open subset U of X, A / U !=
{}.> I have seen two definitions of 'separable topological
space':> a) (X,T) is separable if there exists A <subset> X,
where A is countable and> dense in X> b) (X,T) is separable if
there exists A <proper subset> X, where A is> countable and
dense in X> Given def (a), its simple to prove that all
countable spaces X are> automatically separable since the
subset X is countable and non-trivially> intersects every
nonmpty open subset. However, this won't work given def> (b).
I first became suspicious when I saw a proof that all
countable spaces> were separable that seemed ridiculously
complex in comparison to the> seemingly obvious 1 step proof
above. I later found definitions of> separable like def (b).>
l8r, Mike N. ChristoffI have never seen definition (b). I have
seen the symbol $subset$used to mean subset, not proper subset,
so maybe that is what you saw.Certainly a one-point space with
the discrete topology is to beconsidered separable, even
though it has no dense proper subsets.--
http://www.math.ohio-state.edu/~edgar/===
===
Subject:
: Re: How many
different resistances with n resistors?X-ID:
E2viu0ZC8eLZuWnmdUrIGOhhK5Qw9k9v9kIYQ8-4TI9+4-xlZWn+kbtchow@
lsa.umich.edu schrieb:>tchow@lsa.umich.edu schrieb:>The Puzzle
Corner in MIT's magazine Technology Review raised the>question
of the number of distinct resistances achievable by
connecting>ten unit resistors.>Tony oletti (seq. A048211)
considers only cases that can be reduced by>applying the
formula for parallel/series. If you have 6 resistors there is
1>configuration where this is not possible. I suppose we get
an additional>resistance value.> What's the 6-resistor
configuration you have in mind?An H with asymmetric legs. I
thought that 6 is the smallest number with thisproperty, but
obviously a H with 5 resistors gives us r=1 which seems
impossiblewith other combinations of 5 resistors. I did not
yet check, whether the 6-H hasa resistance not occuring among
the (other) 53 possibilities. I will play withthis a little
tomorrow. I think we would find a reference in encyclopedia
ifsomeone had solved the general case.Klaus> And has anyone
calculated the answer to the problem if we don't assume>
series/parallel circuits?> --> Tim Chow
tchow-at-alum-dot-mit-dotdu> The range of our projectiles--ven
... the artillery---however great, will> never exceed four of
those miles of which as many thousand separate us from> the
center of the earth. ---Galileo, Dialogues Concerning Two New
Sciences===
===
Subject:
: Re: Help needed - primes programWell how
about
this:---------------------------------------------------------
----------------------Program PB;Var x,y:word; f:word;
i,n:word; ok:boolean;Begin Write(Enter X and Y -> ');
Readln(x,y); For i:=x to y do Begin f:=3; ok:=true; If (i=1)
or (i mod 2 = 0) then Begin f:=i+1; ok:=false; End; While
f<=sqrt(i) do Begin If i mod f = 0 then Begin f:=i; ok:=false;
End else f:=f+2; End; If ok=true then Write(i,' is a prime. ');
End;
Readln;End.---------------------------------------------------
----------------------------It ignores all number with form
k*2 , k{2,3,4,...} and it goes till sqrt(i)insted of i.Is this
faste or slower than sieve of eEatosthenes? Why?===
===
Subject:
: Re:
A newbie's question -- about real number> Two real numbers r1 =
0.89, r2 = 0.889999999..... (with infinite 9s), > can we say r1
= r2 ?Yes. The problem is that we present infinite decimal
expansions in school before their meaning can be properly
understood. Once this meaning is made clear, proving r1 = r2
is a very simple exercise.===
===
Subject:
: Re: Teaching philosophyI
promise to cheerfully accept students who don't have the>
prerequisites and can't add fractions without a calculator or
solve> linear equations. I will have them work in groups so,
hopefully,> someone in each group can do the work. I will be
their best friend so> they don't think of me as a teacher. We
will have nice pleasant> classes where we all discover things
together so nobody feels left out> or has their selfsteem
bruised. Before each exam I will give out a> practice exam
****very**** similar to the real exam so they will know> what
to expect. I will be very generous with part credit for those>
that still don't get it, even giving re-takes if necessary. I>
**guarantee** a very high success rate in my classes. There
will be so> many A's and B's that I will get awesome student
evaluations. Students> will flock to my classes because I am
such a great teacher. The Dean> will notice the improvement in
my department's success rate and will> probably increase it's
funding. Over time, as my methods catch on,> even the
University administration and maybe even the Regents will>
hold our department up as a shining example of what is being
done to> improve higher education.Hah, pretty good!===
===
Subject:
:
Re: Math Too Advanced For Mainstream EconomistsRobert Vienneau
<rvien@see.sig.com>threatening to hold his breath until he
dies once again,> The pitiful child who posts below is teaches
economics at a leading> second-tier department of economics in
the U.S.A. Hard to believe.Maybe you can get your big brother
to beat him up.===
===
Subject:
: Re: help with solutions for three
questions from the past contest> 3. A boat with an ill
passenger is 7.5 mi north of a straight> coastline which runs
east and west. A hospital on the coast is 60> miles from the
point on shore south of the boat. If the boat starts> toward
shore at 15 mph at the same time an ambulance leaves the>
hospital at 60 mph and meets the ambulance, what is the total
distance> (to the nearest 0.5 mile) traveled by the boatand
the ambulance? ans:> 62.5> Boat . | | 7.5| z | Shore
|___________________________________.Hospital x 60-xHours
along distance z at 15 mph must equal hours along distance60-x
at 60 mph.And at constant speeds, distance = speed * time.>
I'll be curious if that's the best solution. (It may be the
one expected.)> A better solution would be to minimize the
time to the hospital. This does minimize the time to the
hospital!===
===
Subject:
: Re: question about periodic function>If
you have two periodic functions f(t) and g(t) with periods a
and b,>then the function f(t) + g(t) is periodic with period c
where c is the>least-common-multiple of a and b. Here the first
period is 1/.018=500/9 nd>the second period is 1/.02=50. So the
least-common-multiple is 500, i.e.>1/.002. That should be the
period of the combined function.> How to prove the theorem?
Any hint is welcome. TIA.The sum will only be periodic if the
ratio of periods is rational.If that ratio is irrational f(t)
+ g(t) is not periodic.===
===
Subject:
: Re: Simple numbers> Well
what is that algorithm?It's called the Sieve of
Eratosthenes.-- Dave SeamanJudge Yohn's mistakes revealed in
Mumia Abu-Jamal
ruling.<http://www.commoncouragepress.com/index.cfm?action=
book&bookid=228>===
===
Subject:
: Anyone knows a formula for this?
(combinatorics)HiLet n be the number of cards drawn from an
ordinary deck (52 cards).For example according to the book, if
n=5i.e if we pick 5 cards randomly from a deck of 52 cardsThe
probability of getting a) A full house
is:[C(4,2)*C(4,3)*2*C(13,2)]/C(52,5)b) Exactly two pairs
is:[C(4,2)*C(4,2)*C(44,1)*C(13,2)]/C(52,5)So i was wondering
if there's a general result (a formula)? for a) and b)where n
is the number of cards drawn.===
===
Subject:
: Re: errors in an
argument> If I were trying to explain the error to someone, I
would probably> use the analogy of rolling a 20 sided dice 100
times and recording> the result. By the proof above, one can't
do this, since the> probability of any particular sequence is
(essentially) 0.What about the probability of any particular
sperm actually being the oneto fertilize an egg? That would
seem to preclude the possibility of sexualreproduction
altogether.-- Joe Bramblett, KD5NRH===
===
Subject:
: Re: I'd like to
join this group!> Does the Equation below really prove that 2=1
?? Y N > 1) X=Y ; Given> 2) X^2=XY ; Multiply both sides by X>
3) X^2-Y^2=XY-Y^2 ; Subtract Y^2 from both sides> 4)
(X+Y)(X-Y)=Y(X-Y) ; Factor> 5) X+Y=Y ; Cancel out (X-Y) term>
6) 2Y=Y ; Substitute X for Y, by equation 1> 7) 2=1 ; Divide
both sides by Y> Go back to step 4. What it really proves is
2x0=1x0. > Now try this:> 1) I am nothing without her.
(Shakespeare)> 2) ME - SHE = 0 (Algebraic representation)> 3)
ME = SHE (Added SHE to both sides)> 4) M = SH (Divided by E)>
5) MIT =  (Multiplied by IT)> So there you have the
mathematical proof of what Shakespeare > thought of the future
American techincal education. :-)You Guys are too sharp.That
was actually printed in OMNI magazine Nov 79 , as proof that
2=1 . ;-)===
===
Subject:
: Re: Dik Winter's claims revisited,
dependency issue> James Harris:> |> They've gone so far that
Keith Ramsay even claimed to have posted a> |> solution for
the w's, but how did he pick w's from infinity?> I'm not sure
what you mean here by pick w's from infinity.> | He didn't.
There is a well-defined algorithm for computing> |the w's,
starting with the number (1 + sqrt(-167))/2. If you> |had
replaced 7 with 17, the starting number would have been> |(-1
+ sqrt(-407))/2 and the result would be different.> Let's
recall what I posted.> |It appears that Q(sqrt(-167)) has a
class group of order 11.> |Let r stand for (1+sqrt(-167))/2
(which is an algebraic integer,> |in spite of the 2 in the
denominator).> |> |r^11 = (-592764018-86559857*r)> | =
(44555-222*r) (-12882-2017*r)> |> |and> |> |7^11 =
(44555-222*r)(44555+222*r).> As an intermediate step, I found
the GCD of ((1+sqrt(-167))/2)^11> and 7^11, which is an 11-th
power of the GCD of (1+sqrt(-167))/2> and 7.> Harris to check
the result using a calculator or something like that.> |Also
(-12882-2017*r)(-12882-2017*(1-r)) = 2^11 * 3^11, which is>
|relatively prime to 7^11.> |> |The factor 44555-222*r = 44444
- 111 sqrt(-167) and its conjugate> |44444 + 111 sqrt(-167) are
roots of x^2-88888*x+7^11=0. One GCD of> |r with 7 is then
(44444-111 sqrt(-167))^(1/11), which is a root of> |x^22 -
88888 * x^11 + 7^11 = 0.All those numbers are hard to parse
through, so I just went back toconsidering thata_1(x) + a_2(x)
= (x-1)so assuming factors f_1(x), and f_2(x), with yourw_1(2)
= (44444-111 sqrt(-167))^(1/11), andw_2(x) = (44444+111
sqrt(-167))^(1/11), I have w_1(2) f_1(2) + w_2(2) f_2(2) =
1and assuming the w's are symmetric by sign, I have that
f_1(2) mustequal f_2(2).But f_1(2)f_2(2) = 6as the problem I
see is that you're stuck, if the w's are symmetric bysign, so
are you saying they are not?===
===
Subject:
: Re: How many ways to
put 5 balls into 500 ordered cups?> You are given 500 numbered
cups and five identical balls. Any cup can> hold up to five
balls. How many ways can you put the five balls into> the 500
cups?Three; you can use your left hand, your right hand, or
both. Of course,you could use some other sort of mechanical
aid, but that would be aquestion for the engineering groups.--
Joe Bramblett, KD5NRH===
===
Subject:
: Re: A newbie's question --
about real number> Hence the rationals are uncountable! >Nice
try, but no cigar. The set of decimal sequences that terminate
in all >0's or all 9's is countable.No cigar? Bummer. But I
*can* find an uncountable number of
non-terminatingrepresentations of 89/100 and figured that
since r1<>r2 the conclusionfollowed. If that is not true then
clearly 89/100 is irrational! Even better.===
===
Subject:
: Re: No
Set Contains Every Computable Natural> Russell Easterly
says...> The usual notion of recursive set is that a set S
is> recursive as a *subset* of the naturals if there is a>
Turing machine T such that for any *natural* number n> (in
unary notation, to be specific)>The set of all natural
numbers is not a proper subset of>the set of all natural
numbers.> But it *is* a subset. A recursive subset.> T(n)
halts and outputs 1 <-> n is an element of S> T(n) halts and
outputs 0 <-> n is not an element of S> If x is some input
that is not a representation of a natural> number, then there
is no constraint on what T(n) does.>According to this
definition, the set of all natural numbers>is recursive
because there exists a TM that doesn't even>read the input and
always outputs a 1.> Right.The set of all natural numbers is
recursive by fiat.> No, not by fiat. It is a trivial
consequence of a general definition--> the definition that
Daryl McCullough has reproduced for you.Using Daryl
McCullough's definition, I can claim the set ofall natural
numbers is (1,2,3). Since the string 1111 doesnot represent a
natural number in my system, it is too largeto be the input to
a Turing machine.Why bother to include computability in the
definition?> There is no separate definition for the case of
all natural numbers.> There is one definition that applies to
all subsets of the set of natural> numbers. The set of all
natural numbers (henceforth N) is a subset of> N. We apply the
defintition to N and find out trivially that it is> recursive.>
It is not a trivial definition. It is quite profound. But it
is> trivial that N is recursive.I find the idea that N=(1,2,3)
quite profound.A human must decide if x is a member of this
setbefore x can be given to the TM.> You are seeking to create
a new definition and theory of computability.> Good. You want
to consider a larger context of inputs other than> just
natural numbers. For example, you want to include infinite>
stings as acceptable inputs.Who decides the string is finite?A
TM will always think the input is finite.I have given proofs of
this.> Fine. I would agree with you that> if we enlarged the
universe of acceptable inputs from just N to a larger> set
containing infinite strings, then N would probably not be>
recursive in any reasonable new definition of the word. You're
right,> a TM would not be able to conclude that an infinite
string of 1's is> not a natural number.A TM will say that any
string of 1's represents a natural number.> I would say that
in this enlarged context, that> N would be recursively
enumerable but not recursive.> But your problem is that you
are using terms with agreed upon definitions> by the
mathematical community (like recursive) incorrectly. You have>
your own personal definition that you have failed to state.I
used a standard definition of decidable language.The standard
definition states that a human must decidethat a string is
finite before giving it to the TM.Ultimately, a human being
must decide if a string, x,represents a natural number.Why
pretend there is an algorithm that can do this?> This is why>
mathematicians are arguing with you and pointing out your
obvious> errors. If you would like to re-define recursive by
all means do so.I am not redefining recursive.I am pointing
out that if the definition of decidable languangesays there is
a TM that can decide if string, x, is a memberof the language,
then the language of all unary representationsof the natural
numbers is NOT a decidable language.If we include the
requirement that a human being decideswhether x is a finite
string, then I agree N is recursive.This additional
requirement allows me to choose N to beany set I want it to
be.> But make your new definition known. It would also be good
to give> it another name to avoid confusion, like
Easterly-Recursive.> Until you do this, you will probably
continue to make false statements> that are easily refuted.Are
you refuting my claim that N=(1,2,3)?> Remember, mathematics is
precise.N is precisely equal to (1,2,3).> By the way there are
already in existence more generalized recursiontheories,>
which deal with recursion on other domains besides N.Can you
give references?Obviously, I need to learn more about this
stuff.Russell- 2 many 2 count===
===
Subject:
: Re: Anyone knows a
formula for this? (combinatorics) Adjunct Assistant Professor
at the University of Montana.>Hi>Let n be the number of cards
drawn from an ordinary deck (52 cards).>For example according
to the book, if n=5>i.e if we pick 5 cards randomly from a
deck of 52 cards>The probability of getting >a) A full house
is:>[C(4,2)*C(4,3)*2*C(13,2)]/C(52,5)>b) Exactly two pairs
is:>[C(4,2)*C(4,2)*C(44,1)*C(13,2)]/C(52,5)>So i was wondering
if there's a general result (a formula)? for a) and b)>where n
is the number of cards drawn.You should try to understand what
the numbers mean.For example, with full house, you have
C(52,5): that's the number ofdifferent ways in which you can
randomly pick 5 cards from a deck of52 cards. It is the total
space. If you are randomly pickin k cardsfrom a deck of 52,
then that number should obviously be replaced byC(52,k).What
do the other numbers mean? I.e., C(4,2)*C(4,3)*2*C(13,2)should
be the number of different 5 card hands which are a fullhouse.
How is this accomplished? Why does this count that?Well, a
full house is determined by stating the following things: (1)
The rank of the 3-of-a-kind (i.e., is it 3 aces, 3 jacks, 3
tens, etc); and (2) The rank of the pair; and (3) The suits
involved in the 3-of-a-kind; and (4) The suits involved in the
pair.That completely determines the hand, since there are no
cards leftoverin a 5-card hand.Now, C(4,2) means how many ways
to choose 2 cards out of 4; thatrepresents (4), how to pick
which suits the pair will involve.C(4,3) is how you pick the
suits for the 3-of-a-kind.And C(13,2) picks the two ranks that
will be involved; you multiply by2 because it matters which
goes first (i.e., is it tens over fours, orfours over tens)?
So, say you are drawing k cards instead, k>5, and you want to
know howmany ways in which you can have exactly a full house.
Obviously, somevalues of k will not allow it, so it becomes
more complicated (if youdraw 50 cards, say, then you will
surely have a better hand than afull house to pick out of
there). But you determine the full house thesame way. What
remains is to ensure that the left-over cards in yourhand do
not add more to your hand. So, for example, you would need
tochoose (k-5) more cards, making sure that none of them are
in the samerank as the pair or as the 3-of-a-kind; you would
also have to ensurethat they are not pairs among themselves,
say, that you do not drawenough to get a 5-card flush,
etc.Some easy ones, though: say we have k=7; there's no danger
of a flushor a straight, so all you have to do is make sure you
do not draw fromthe same rank as the pair or the 3-of-a-kind,
and that the two extracards are not a pair. So you need to
pick 2 cards of different ranks;there are 11 ranks left, so
you multiply by C(11,2). And you need toselect their suits.
Pick one suit for the higher card, C(4,1), andanother suit for
the other card, C(4,1). So we get that, drawing 7cards, the
number of hands that have exactly a full house and nobetter
(and do not have a 3-of-a-kind and two pairs)
are:C(4,2)*C(4,3)*2*C(13,2)*C(11,2)*C(4,1)*C(4,1)out of a
total of C(52,7) possible hands.So the probability would
beC(4,2)*C(4,3)*2*C(13,2)*C(11,2)*C(4,1)*C(4,1)/
C(52,7).Higher number of cards complicates matters, since you
would have todiscard hands that get a straight, a flush,
etc.-- =====Arturo Magidinmagidin@math.berkeley.edu===
===
Subject:
:
RFI:Wanless' Fourth Conjecture/Dirichlet's Geometric TheoremThe
series a^n+b [hcf(a,b)=1] includes an infinite number of primes
[overall n]===
===
Subject:
: Re: RFI:Wanless' Fourth
Conjecture/Dirichlet's Geometric TheoremImmediate
proviso:a>1,but apart from that...?> The series a^n+b
[hcf(a,b)=1] includes an infinite number of primes [over> all
n]===
===
Subject:
: Superfluous information; was: Re: Mean Value
Theorem>Does anyone know of a case where a textbook purposely
included >superfluous infomation as part of an exercise? Or
how about on a >test question? It seems that it would make the
problems somewhat >more realistic, albeit esthetically
unpleasing. It may make students>think more about the problem.
Students seem to develop the unhealthy >assumption that if it
is given, then is must be used.I have often put superfluous
information in homework and exams (mainly in operations
research and statistics) for just this reason.-- Stephen J.
Herschkorn herschko@rutcor.rutgers.edu===
===
Subject:
: Re: A
newbie's question -- easier way> Two real numbers r1 = 0.89,
r2 = 0.889999999..... (with infinite 9s),Hello! I think an
easier way to show the equality is as follows:r1 = .89 = 8/10
+ 9/100r2 = .889999... = 8/10 + 8/100 + 9/900r1 - r2 = 9/100 -
8/100 - 9/900 = 9/100 - 81/900 = 0.0Skip===
===
Subject:
: Re: No Set
Contains Every Computable Natural>and hence derive a notion of
computability, recursive>sets, etc. These notions as stated
make the claim that N is recursive>trivial.Not completely
trivial. We are letting a human decide what is computable.N is
a recursive set by fiat.> I think that's a fair assessment.> We
have the mathematical object N. We have the (also
mathematical)> notion of Turing machines which produce marks
on tapes. If we want to> reason about computability of
functions involving the former using the> latter, then we must
choose a representation of natural numbers by> tapes.> If we
ask whether N is recursive, then it depends on what context
we> mean. We may mean whether there is a computable function N
-> N which> returns 1 iff its input is in N and 0 else. To
prove that there is> such a function, we need to reason about
Turing machines, but only> about their behavior on input tapes
which represent natural numbers.Ultimately, a human decides
what represents a natural number.Computability has nothing to
do with it.> On the other hand, we might ask whether there is
a total function> which, on *any* input tape whatsoever,
returns 1 iff the tape> represents a natural number and 0
else. Now, if any includes tapes> which have infinitely many
ones at the start, then clearly the answer> is no. In *this*
sense, N is not recursive.This is what I was trying to say.>
Again, there is a difference between the following two
questions:> (1) Does there exist a TM t such that, for every
input tape i which> represents some natural number n, t(i) =
1?Trivially, yes.1) Print 12) Halt> (2) Does there exist a TM
t such that, on every input tape (including> those with an
infinite number of ones) i, t halts and outputs 1 iff i>
represents some natural number?No TM can do this and halt
after a finite number of steps.Why bother to pretend N is
computable?> I don't know what it means for N to be
computable. I think it's> pretty clear what I mean when I say
N is a recursive set. Evidently,> it's not clear to you, but
maybe the above helps.Given any set, S, of representations of
natural numbers,I can write a TM that will find a
representation of a natural numberthat is not in S. (No, this
TM does not halt after a finite numberof steps.)Obviously, a
human has to decide if a symbol representsa natural number. No
TM can do this.> No TM can devise a convention, if that's what
you mean. At least, no> TM can devise a convention in the
sense that matters here.But, a TM can extend the convention in
such a way that no setcontains every symbol that represents a
natural number.A TM doesn't have a well-defined notion of N.>
I wouldn't want to claim that a TM has any notions at
all.Exactly.Computers do what you tell them to do,not what you
want them to do.Russell- 2 many 2 count===
===
Subject:
: Re: Dik
Winter's claims revisited, dependency issue> James Harris:> |>
They've gone so far that Keith Ramsay even claimed to have
posted a> |> solution for the w's, but how did he pick w's
from infinity?> I'm not sure what you mean here by pick w's
from infinity.> | He didn't. There is a well-defined algorithm
for computing> |the w's, starting with the number (1 +
sqrt(-167))/2. If you> |had replaced 7 with 17, the starting
number would have been> |(-1 + sqrt(-407))/2 and the result
would be different.> Let's recall what I posted.> |It appears
that Q(sqrt(-167)) has a class group of order 11.> |Let r
stand for (1+sqrt(-167))/2 (which is an algebraic integer,>
|in spite of the 2 in the denominator).> |> |r^11 =
(-592764018-86559857*r)> | = (44555-222*r) (-12882-2017*r)> |>
|and> |> |7^11 = (44555-222*r)(44555+222*r).> As an
intermediate step, I found the GCD of ((1+sqrt(-167))/2)^11>
and 7^11, which is an 11-th power of the GCD of
(1+sqrt(-167))/2> and 7.> Harris to check the result using a
calculator or something like that.> |Also
(-12882-2017*r)(-12882-2017*(1-r)) = 2^11 * 3^11, which is>
|relatively prime to 7^11.> |> |The factor 44555-222*r = 44444
- 111 sqrt(-167) and its conjugate> |44444 + 111 sqrt(-167) are
roots of x^2-88888*x+7^11=0. One GCD of> |r with 7 is then
(44444-111 sqrt(-167))^(1/11), which is a root of> |x^22 -
88888 * x^11 + 7^11 = 0.OOPS! I'm re-thinking my previous
post. I'll consider Keith Ramsay'spoints here some more.For
all I know I'm wrong, but then again, maybe I'm still just
missingsomething.===
===
Subject:
: Re: I'd like to join this group!
charset=Windows-1252> That was actually printed in OMNI
magazine Nov 79 , as proof that 2=1 .It was also printed in
the Wall Street Journal to explain how companyearnings are up.
;-)-- Growing old is mandatory.Growing up is
optional.===
===
Subject:
: Re: RFI:Wanless' Fourth
Conjecture/Dirichlet's Geometric Theorem_If_ it's true - can
one prove it from Dirichlet's [Arithmetic ie
standard]Theorem?J> The series a^n+b [hcf(a,b)=1] includes an
infinite number of primes [over> all n]===
===
Subject:
: infinitely
many palindromic primes?On page 228 of Albert Beiler's
_Recreations in the Theory of Numbers_(Dover, 1964) one finds
the raw assertion that there are infinitelymany palindromic
primes, primes whose base 10 representation reads thesame
forwards and backwards.The result is certainly believable,
e.g., as a standard heuristicpredicts infinitely many primes
consisting solely of 1's, but I wassurprised to see it claimed
as a theorem. Is this really the case? Anold (1984) sci.math
post from Bob Silverman suggests that it is andthat the proof
is connected with results on automorphs of binaryquadratic
forms. (The post in question discusses classes ofautomorphic
binary quadratic forms, but the references to Dicksonseem to
be to theorems about automorphs of binary forms and
theirconnection with the Pell equation.) Can anyone elucidate
for me theconnection of this theory to palindromic
primes?Paul===
===
Subject:
: Metric Enginer The Big Rip?This is
expanded versionNew 269 pageSuper Cosmosnow
athttp://qedcorp.com/destiny/SuperCosmos.pdfThe Hunt for the
Zero Point
Farcehttp://www.salon.com/books/review/2002/08/05/zero_gravity
/May The Farce be not with you.FYI draft in progress with more
re: ISSO/Sarfatti etc.DARK MATTERS SURROUND DARK ENERGYTwo big
stories from the world of physics may portend the arrival of
new weapons of mass destruction far more powerful and compact
than atomic bombs. In recent years it has been discovered that
our universe is being blown apart by a mysterious anti-gravity
effect called dark energy. Mainstream physicists are
scrambling to explain this mysterious acceleration in the
expansion of the universe. Some physicists even believe that
the expansion will lead to The Big Rip when all of the matter
in the universe is torn asunder - from clusters of galaxies in
appears to be made of two unknowns - roughly 23% is dark
matter, an invisible source of gravity, and roughly 73% is
dark energy, an invisible anti-gravity force. Ordinary matter
constitutes perhaps 4 percent of the universe. Recently the
British science news journal New Scientist revealed that the
American military is pursuing new types of exotic bombs -
including a new class of isomeric gamma ray weapons.That was
an original idea of mine in 1963 at Cornell and I discussed it
with Hans Bethe. That is one of the reasons Ron Bullough
invited me to Harwell in 1966. No doubt others thought of it
but probably later. I thought of it while at Tech/Ops in
Lexington, Mass working for George Parrant Jr.Unlike
conventional atomic and hydrogen bombs, the new weapons would
trigger the release energy by absorbing radiation, and respond
by remitting a far more powerful radiation. In this new
category of gamma-ray weapons, a nuclear isomer absorbs x-rays
and remits higher frequency gamma rays. The emitted gamma
radiation has been reported to release 60 times the energy of
the x-rays that trigger the effect.Gamma-ray weapons could
trigger next arms race 19:00 13 August 03Exclusive from New
Scientist Print Edition. Subscribe and get 4 free issues.An
exotic kind of nuclear explosive being developed by the US
Department of Defense could blur the critical distinction
between conventional and nuclear weapons. The work has also
raised fears that weapons based on this technology could
trigger the next arms race. The explosive works by stimulating
the release of energy from the nuclei of certain elements but
does not involve nuclear fission or fusion. The energy,
emitted as gamma radiation, is thousands of times greater than
that from conventional chemical explosives. The technology has
already been included in the Department of Defense's
Militarily Critical Technologies List, which says: Such
extraordinary energy density has the potential to
revolutionise all aspects of warfare. Scientists have known
for many years that the nuclei of some elements, such as
hafnium, can exist in a highnergy state, or nuclear isomer,
that slowly decays to a lownergy state by emitting gamma rays.
For example, hafnium-178m2, the excited, isomeric form of
hafnium-178, has a half-life of 31 years. The possibility that
this process could be explosive was discovered when Carl
Collins and colleagues at the University of Texas at Dallas
demonstrated that they could artificially trigger the decay of
the hafnium isomer by bombarding it with lownergy X-rays (New
Scientist print edition, 3 July 1999). The experiment released
60 times as much energy as was put in, and in theory a much
greater energy release could be
achieved.http://www.newscientist.com/news/news.jsp?id=
ns99994049I was thinking in 1963 of a gamma ray laser pumping
a nuclear isomeric transition. Bethe at the time said it
wouldn't work and basically discouraged me working on
it.Bekkum continued:In the summer of 2000 I contacted Nick
Cook, the former aviation editor and aerospace consultant to
Jane's Defence Weekly, the international military affairs
journal. Cook had been investigating black budget super-secret
research into exotic physics for advanced propulsion
technologies.Uh Oh :)I had been monitoring electronic
discussions between various American and Russian scientists
theorizing about rectifying the quantum vacuum for advanced
space drive. Several groups of scientists, partitioned into
various research organizations, were exploring what NASA calls
Breakthrough Propulsion Physics - exotic technologies for
advanced space travel to traverse the vast distances between
stars. Partly inspired by the pulp science fiction stories of
their youth, and partly by recent reports of multiple radar
tracking tapes of unidentified objects performing impossible
maneuvers in the sky, these scientists were on a quest to
uncover the most likely new physics for star travel. The NASA
program was run by Marc Millis, under the Advanced Space
Transportation Program Office (ASTP). Joe Firmage, a Silicon
Valley entrepreneur, who at age 28 had found risen to CEO of a
three billion dollar internet firm, began to fund research in
parallel with NASA. He hired a NASA Ames nanotechnology
scientist, Creon Levit, to run the International Space
Sciences Organization,Joe did that because I suggested it. I
introduced Creon to Joe.Cook was intrigued by the apparent
connections between various private investors, defense
contractors, NASA, INSCOM (American military intelligence),
and the CIA. While researching exotic propulsion technologies
Cook had heard rumors of a new kind of weapon, a sub-quantum
atomic bomb, being whispered about in the dark halls of
defense research.I think that must have come from me regarding
J. P. Vigier's tight atomic states with experiments in Beograd,
Serbia by Z. Maric (same name as Einsteins wife, Mileva Maric)
and G. Dragic. But how did Cook hear about that? We brought
Vigier to ISSO in San Francisco several times along with
physicist Gennady Shipov from Moscow. That story with
photographs of Vigier and the group is in my autobiography
Destiny Matrix. Dragic A, Maric Z, Vigier JP; Phys. Lett. A
265 (2000) 163. New quantum mechanical tight bound states and
'cold fusion'. Creon Levit and Vigier met with Maric in
Budapest, Hungary in 2000.Bekkum who is one of my ISEP on line
virtual students continued:Sub-quantum physics is a
controversial re-interpretation of quantum theory, based on
so-called pilot wave theories, where an information that the
predictions of ordinary quantum mechanics could be recast into
a pilot wave information theory. Recently Anthony Valentini of
the Perimeter Institute has suggested that ordinary quantum
theory may be a special case of pilot wave theories, leaving
open the possibility of new and exotic post-quantum
technologies. Even thought rumors of a sub-quantum bomb may be
purely fantasy It's not fantasy. It might not work, Maric and
Dragic in Beograd, while not ostensibly trying to make a
weapon by any means, were trying to test Vigier's basic theory
of the spatially extended electron, which I think is basically
a correct idea and fits my own ideas including why the
electron appears to shrink to less than 10-16 cm under high
resolution imaging (i.e. scattering) and how the electric
charge distribution is contained by the strongly attractive
short-range zero point stressnergy density w = -1 exotic
vacuum tiny dark matter core only roughly one Fermi (10-13 cm)
across. This solves the Abraham-Becker-Lorentz-Poincare
compensating stress problem of 100 years ago. My new
explanation of this long-standing mystery that stumped even
Richard Feynman, only works in the Bohm hidden or extra
variable interpretation. That is, a classical spatially
extended electric charge distribution is unstable. It explodes
under its own self-repulsion. This is why physicists had to
postulate a point electron because they did not understand
that the strong short-range gravity attraction of the positive
zero point pressure in the dark matter phase of exotic vacuum
would hold the charge together. As Herbert Frohlich told me at
UCSD in La Jolla in 1966 the basic thing wrong with physics is
the idea of the point electron. The bad idea of the point
electron gives the infinite energy in quantum electrodynamics.
Richard Feynman told me in his office at Cal Tech in 1968 that
infinite renormalization is a bait-and-switch shell game like
shoddy used car salesmen use, and like the lame WMD excuse to
liberate Iraq, and it is a scandal in physics that no one
could do better than what he had done. No wonder all our top
theoretical physicists did not balk at sharing Werner Erhards
fine Cuban cigars and liquour in his attic at The Brown House
on Franklin Street in San Francisco. (Grin) Physicists did not
know 100 years ago that 1/3 or so of the universe was this kind
of exotic vacuum. For example, there is a huge sphere (AKA
galactic halo) of exotic vacuum of w = -1 positive pressure
that holds our galaxy together preventing our solar system
from going off into space on its own. This sphere looks like w
= 0 cold dark matter from our vantage point. What works on this
large scale also works on the small scale of the single
electron (and all the charged lepto-quarks). A neutrino has
some mass and is simply a micro-geon of Mass without mass pure
zero point energy with positive pressure. One can plausibly
metric engineer exotic vacua into artificial black holes and
traversable wormhole star gates as well as weightless warp
drives as we allegedly see in the seemingly impossible Wheeler
wormhole with at least two mouths (like 2D spherical surfaces
in 3D space) and quantized electro-weak-strong Faraday lines
of gauge force forming closed loops with no free ends
threading the mouths. The fact that there is very little
naturally occurring anti-matter in our universe is direct
evidence for at least one parallel universe next door where
all the twin wormhole mouths are attached to. This is
consistent with Jacques Vallees Fastwalker. Furthermore, none
of the dark matter detector experiments now planned or
underway will ever click with the right stuff if my ideas here
are correct. This is a matter of fundamental principle. there
is no question that physicists seriously contemplate a phase
transition in the quantum vacuum as a real possibility. The
quantum vacuum defies common sense, because empty space in
quantum field theory appear and disappear far too quickly to
be detected directly, but their existence has been confirmed
by experiments that demonstrate their influence on ordinary
matter.A major component of the physical quantum vacuum
consists of virtual electrons frothing and bubbling, like The
Brew of the Three Witches in Shakespeares Macbeth, at the
foamy Fermi surface edge of the Dirac negative energy sea.
This is because of the Pauli exclusion principle that only
none or one electron per quantum state. A virtual electron
pops out of the vacuums Fermi surface leaving a hole behind.
The hole is the virtual positron. The result is a virtual
electron-positron pair. However, the virtual electron and the
virtual positron attract because they have opposite charges
and they are exchanging virtual photons. Therefore, some of
them form a more stable bound state. An enormous number of
these virtual pairs Boseinstein condense into the same center
of mass quantum wave packet of the single pair forming the
Vacuum Coherence Field (AKA Inflation Field). This is a
dynamic steady state nonquilibrium dissipative structure of
detailed balance in which there is a continual inflow and
outflow of virtual pairs into and out of this giant quantum or
macro-quantum superfluid. That is, virtual pair bound states
are continually created and destroyed from and into the
electrically neutral ionized plasma of broken virtual pairs
which is the zero point dark energy/matter normal fluid
component of the exotic vacuum that sustains the robust
generalized phase rigidity of the macro-quantum vacuum
coherence field. This P.W. Anderson phase rigidity explains
Andrei Sakharovs metric elasticity or space-time stiffness or
string tension. Essentially this is a vacuum phase transition,
similar to the BCS phase transition from a normal metal to an
electrical superconductor, from the globally flat
micro-quantum electrodynamic vacuum without any gravity at all
to the curved macro-quantum electrodynamic vacuum with emergent
gravity. Einsteins field equation of general relativity can be
derived from the phase wiggles and ripples in the robust
stable macroscopically occupied center of mass quantum wave
packet of the bound state of the virtual electron-positron
pair. The exotic vacuum dark energy and dark matter are simply
the amplitude wiggles and ripples of this same virtual pair
quantum wave packet. The wave packet spreads over the entire
3D space of the post-inflationary bubble on which our
Hubble-horizoned universe is located along with an infinity of
parallel universes next door as Max world hologram idea is
correct, take the surface area of the expanding Hubble sphere
that is the causal retarded boundary of 3D space of our past
light cone at Earth and divide it by the quantum of area. That
gives us the number of Bekenstein-Shannon c-bits and explains
the irreversible arrow of time (AKA Second Law of
Thermodynamics) of increasing thermodynamic entropy in terms
of the dynamical expansion of the 3D space of the universe.
Lenny Susskind calls this kind of idea DeSitter Space, which
is a large-scale homogeneous isotropic limiting case of what I
am pointing to on all scales.As Above So BelowSuch research
should be forbidden!Sakharov to Lev OkunTiger TigerIts already
too late. Pandoras Box is open. Schrodingers Cat has jumped out
of it. This Kitty has grown up to William Blakes Tyger, Tyger
shining bright in the darkness of the night and I am holding
it by the tail. Hold on to your hats for a bumpy ride until we
switch on zero g-force weightless warp drive!Tiger, tiger,
burning brightIn the forests of the night,What immortal hand
or eyeCould frame thy fearful symmetry?In what distant deeps
or skiesBurnt the fire of thine eyes?On what wings dare he
aspire?What the hand dare seize the fire?And what shoulder and
what artCould twist the sinews of thy heart?And, when thy heart
began to beat,What dread hand and what dread feet?What the
hammer? What the chain?In what furnace was thy brain?What the
anvil? What dread graspDare its deadly terrors clasp?When the
stars threw down their spears,And water`d heaven with their
tears,Did He smile His work to see?Did He who made the lamb
make thee?Tiger, tiger, burning bright,In the forests of the
night,What immortal hand or eyeDare frame thy fearful
symmetry?In the early 1970's Soviet physicists were concerned
that the vacuum of our universe was in fact only one possible
state of empty space. The fundamental state of empty space is
called the true vacuum. Our universe was considered to reside
in a false vacuum, protected from the true vacuum by the wall
of our world. A change from one vacuum state to another is
known as a phase transition. This is analogous to the
transition between frozen and liquid water. Lev Okun, a
Russian physicist and historian recalls Andrei Sakharov, the
father of the Soviet hydrogen bomb, expressing his concern
about research into the phase transitions of the vacuum. If
the wall between the vacuum states was to be breached,
calculations showed that an unstoppable expanding vacuum
bubble would continue to grow until it destroyed our entire
universe! Sakharov declared Such research should be forbidden!
since there was always the possibility that an experiment might
accidentally trigger a vacuum phase transition.British
Astronomer Royal, Sir Martin Rees, Master of Trinity College,
and Director of the Cambridge University Institute of
Theoretical Astronomy on Madingley Road where Stephen Hawking
works discusses all this in Chapter 9 of his important book
Our Final Hour.Could the wall of our universe be breached from
within? The amount of energy required to punch a hole through
the wall appeared to be enormous, and no known natural
physical phenomena, even the most energetic, had punched
through either. A recent report commissioned to examine
potential dangers at the Large Hadron Collider, one of the
next the best of our existing knowledge. Others are not so
certain, however. At least one of the Russian physicists I had
corresponded with was said to have been a former associate of
Andrei Sakharov. He strongly hinted at new theories the
Russians had developed which allow for the manipulation of the
fundamental constants of nature, but he never revealed more
than a sketch of his ideas. He claimed that a breakthrough was
within reach, perhaps within five years Who was that? Not
George Ryazanov?Recent theoretical explorations may suggest
another approach to the physics of the vacuum. The invisible
gravitating dark matter could be the other side of the
invisible dark energy coin, and that suggests the possibility
of manipulating the vacuum for energy release.Now this is my
original idea that you got from our communications over the
past few years. I am the only physicist in the world today, as
far as I know who has suggested this and has already published
it in my two books of 2002 so its in the official record at
the Library of Congress.If a controllable parameter could be
found to mediate the balance between the invisible dark
forces, the result would unleash the vacuum energy of creation
in all of its awful power and majesty. If it were possible to
control the dark sides of the force then spacetime, the arena
where everything we know takes place, could be bent and
twisted with infinitely greater ease than was ever suspected.
This would open Pandora's box to everything from vacuum energy
weapons of mass destruction (capable of destroying the
universe!) to spacetime warp drives and time machines.Exactly,
the above is the thesis of all my books since 2002 at least.A
quick survey of the international electronic archive of
physics papers at www.arXiv.org shows that research into the
vacuum of spacetime for energy production is alive and well.I
do not think that is true. You need to cite specifics here.
There are lots of flaky new age papers on free energy on the
Web written by people without any real credentials but they
are not on www.arXiv.org which is not even allowing competent
fringe papers in controversial topics like cold fusion. So
what exactly are you thinking of here? Indeed, Paul Ginsparg,
who controls the archive, does not even allow Carlos Castro to
publish conservative competent papers on Clifford Algebras
which are not fringe at all!Most authors are independent
researchers struggling with limited funding and resources, yet
their theoretical results suggest that somewhere in Nick Cook's
black world, a major breakthrough has already taken place. Most
likely Israel, UK, the United States and Russia are in the
lead, but China, France, Serbia, Ukraine, Iran, India and
Saudi Arabia all have scientists actively pursuing the
fundamental physics that determine the fabric of our reality,
and are seeking the theory and the means to access the
enormous energies locked inside of the vacuum since the
creation of the universe. Even if the black budget world has
yet to unleash the enormous potential of vacuum energy, there
are signs that those in power may have begun to take notice.
Dr. Harold Puthoff, a scientist with strong government
connections, who has previously worked on classified projects
for the CIA, is a major proponent of vacuum energy physics.
Nick Cook's book, The Hunt for Zero Point, and his recent
stories on zero point energy in Jane's Defence Weekly have
also brought attention to the dangers and military potential
of exotic vacuum research. The American intelligence community
financed so-called psychic spies for over twenty years and
through four presidential administrations. It is highly
unlikely that they would ignore the potential of the quantum
vacuum. Dr. George Chapline, of the Lawrence Livermore
National Laboratory, and Dr. Jack Sarfatti in San Francisco,
knew each other in the sixties in La Jolla, have independently
proposed that the quantum vacuum may unstable to the formation
of coherent virtual processes. Sarfatti suggests that gravity
is an emergent property determined by the physics of the
vacuum. His idea is to find a means of directly interacting
with the physics of the vacuum that controls the shape of
spacetime. Such a possibility would be consistent with the
reported success of Evgeny Podkletnov, the Russian scientist
who is experimenting with spinning superconducting disks.
Podkletnov's most recent papers report the appearance of a
mysterious coherent beam of gravity like radiation with a
measured force of 1000 G. In an interview on BBC radio, Nick
Cook pointed out one immediate application of the Podkletnov
beam - the destruction of missiles and satellites in flight or
in orbit around the earth. Cook showed the BBC internal
documents from Boeing, the American aerospace contractor,
proving their interest in Podkletnov's research.This beam
stuff I am suspicious of. Of course if the experiment is good,
I have to think more about it. I am not so sure if Podkletnovs
experiment is any good and has been replicated.The connections
between Podkletnov's results, and the kind of vacuum research
explored by Sarfatti, beginning in 1999 at the International
Space Sciences Organization are the latest threads in a trail
that most likely originates in cold war disinformation, a game
played by East and West against each other. Glasnost has
shifted the balance of partnerships and the positions of the
players, but not the stakes of an outcome that would leave the
world with even more prolific and powerful weapons of mass
destruction.That is true, as shown in Destiny Matrix, however
you leave out the most important evidence -UFOs!whole business
are the connections. Although Nick Cook never revealed the
identity of his deep throat contact called Dr. Dan Marckus in
the book The Hunt for Zero Point, there was no question that
the Podkletnov results had played a major part in fitting
together the pieces of the puzzle. The amount of interest was
in Podkletnov's reports by NASA, Boeing, and others in the
international arena of aerospace and military research
communities was evidence that there was more here to explore
than the latest musings of the intellectual elite. The truth
is that a fundamental theory of gravity at the scales of
subatomic nuclear physics does not exist. The fact is that no
one understands the nature of the gravitational field at very
small scales. In fact gravity has barely been probed much
below one millimeter. Every attempt to unify the physical
theories of gravity with the well-known standard model physics
of electromagnetism, and the strong and weak nuclear forces,
has failed. More importantly there has been recent progress in
the exotic areas of mainstream research, such as superstring
theory, which suggest new kinds of physics, which might
support explanations for Podkletnov's impulse gravity effect.
One of the current fads in theoretical physics involves large
extra dimensions of space that allow a much stronger version
of gravity to leak off the membrane world of our ordinary
three dimensions. The large dimensional picture allows for the
well known forces of electromagnetism, and the strong and weak
nuclear forces, to be confined to a three dimensional
brane-world floating in a higher dimensional hyperspace.
Gravitons, the quanta of the gravitational force, are viewed
as closed strings, and are able to slip off of our
brane-world, which explains why the gravitational force is so
much weaker than the other forces that hold matter together.
Gravitons could be exchanged between our brane-world and
another brane floating nearby in the same higher
dimensions.The Sarfatti picture offers a more direct
interaction with the new physics than the brane world ideas.
Sarfatti's vision is to find a means of using electromagnetic
fields in the Josephson effect to couple to the virtual
electron-positron pair giant coherent condensate inflation
field inside the vacuum that controls the shape of spacetime
to the real electron pair giant coherent condensate of a
control high temperature superconductor. UCBs Ray Chiao has a
similar idea using a superconductor to transduce
electromagnetic far field waves to gravity waves with high
efficiency conversion. Sarfatti wants to do the same thing
with non-propagating electromagnetic and gravity near fields.
One wonders if the black budget world may have already
produced some of the technology needed to explore and test
these new realms.Not a chance. They are clueless about the
theory. They are still stuck in Hal's PV model and Bernie
Haisch's zero point ideas, which will never, in my opinion
fly. They are not asking the right questions and do not have
the right idea in their minds. It is my belief, until I see
evidence to the contrary, that I am the only physicist on the
planet today who is doing real theoretical work directly
relevant to the achievement of practical metric engineering
anchored in the now observed reality of dark energy. All my
work is public. I would love to be proved wrong on this
especially by Hal Puthoff, but I am not holding my breath.;-)
Extraordinary claims require extraordinary proof. Everything
else I have seen is either on the wrong track asking the wrong
questions like the work of Puthoff, Haisch, Ibison & Rueda for
example, which at least is real physics that has proved itself
wrong in Ibison's PV cosmology paper, or else the claims are
patently obvious nonsense that Feynman called Cargo Cult
Science. There is also the Russian torsion work of Akimov and
Shipov and I am not prepared to make a definitive statement on
that, as the issue is not simple because of several factors
some political. One must be careful there to separate Shipov's
theoretical work from claims made about practical devices
including weapons applications. I do, however, agree with you
that there is a real issue here as defined in Ch. 9 of Martin
Rees's Our Final Hour.Hal Puthoff coined the term metric
engineering for flying saucer technology. Hal has been working
on this problem for many decades and has held high USG security
clearances and has been privy to reliable information that the
saucers are real and are alien. Otherwise he would not be
working on the problem. However, Hal's theories, both of the
zero point energy and of the gravity field will not solve the
problem because they are too naively based and do not ask the
right questions. The basic physics required for this task is
way beyond the depth of Puthoffs self-described engineering
approach and can be found in Rovelli's new book on quantum
gravity.http://www.cpt.univ-mrs.fr/~rovelli/book.pdf===
===
Subject:

: Re: A newbie's question -- about real number> But I *can*
find an uncountable number of non-terminating> representations
of 89/100You can't count to 1?--
http://members.iquest.net/~panoptes/039 53 36 N / 086 11 55
W===
===
Subject:
: Re: any good book on general optimization,
continuous multivariate optimization?> Dear all,> I am
studying general optimization, particularly continuous
multivariate> optimization, something like First Order
Nessecery Condition, Second Order> Nessecary Condition, Second
Order Sufficient Condition for local> minimizers... etc. I got
really confused. The reasons are> 1) Multivariate derivatives,
gradient, etc. are what I am not very familiar> with, so I got
confused;> 2) Some boundary condition and degenerate cases got
me confused: for> example, it is easy to have FONC, SONC, SOSC
for interior points, that's> easy, but what if the FONC, SONC,
SOSC on boundary point, involving the> directional derivative
vector d, etc. And some degenerate cases such as the> Hessian
is indefinite... etc. And the relationship between FONC, SONC,
SOSC,> etc. these all got me confused...> Please point me to
some good books/online lecture notes, etc. that can help> me
clearify these concepts?I recommend you begin by getting an
overview using the excellent book byGill, Murray and Wright.
Itmisses some recent stuff that you probably don'tneed anyway
but otherwise explains everything very well, presenting
themodern point of view (Luenberger is now quite dated)
without highmathematical sophistication. If you don't
understand the math in thereyou'll need to look at a suitable
math text. One cannot get a goodunderstanding of optimization
without becoming familiar with gradientsand Hessians.A good
account of necessary/sufficient conditions is in
Fletcher'sbook. (A comprehensive account is in Mangasarian's
book, but this iswritten for experts.)Arnold
Neumaier===
===
Subject:
: Re: how to understand this statement of
Second order Necessary Condition?> I got confused here because
this seems to me is a half statement:If d'*gradient=0, then
d'*H*d>=0.> but what if d'*gradient>0 but not =0 at that point
x*?In mathematics, if the assumptions are not satisfied,
nothing can beconcluded.> Is this a possible case in the SONC
statment? Is that true that whend'*gradient>0 but not =0,
there is no statement about the SONC?Yes. Yes.Arnold
Neumaier===
===
Subject:
: Re: How to diagonalize a Hermitian matrix>
Hello group,> Could somebody please tell me a numerical
algorithm to> diagonalize a Hermitian matrix H, so that I end
up not only with the> eigenvalues perched along the diagonal,
but also with the unitary> matrix U that conjugates with H to
diagonalize it.> I should probably say what I really want. I
have a> positive-definite Hermitian matrix H, and I need its
matrix square> root. The above is simply the approach that
occurred to me for> obtaining that square-root.> Achava The
group sci.math.num-analysis is a better forum for your
question.And: get hold of Matrix Computations by Gene. G.
Golub and Charles F. Van Loan(newest edition);it has both the
diagonalization and square root algorithms, finely tunedand
analyzed.For ready-made routines for matrices of moderate
size, use MATLAB or itsfree variations (Octave?).Cheers,
ZVK(Slavek)===
===
Subject:
: Re: Dik Winter's claims revisited,
dependency issue> James Harris:> |> They've gone so far that
Keith Ramsay even claimed to have posted a> |> solution for
the w's, but how did he pick w's from infinity?> I'm not sure
what you mean here by pick w's from infinity.> | He didn't.
There is a well-defined algorithm for computing> |the w's,
starting with the number (1 + sqrt(-167))/2. If you> |had
replaced 7 with 17, the starting number would have been> |(-1
+ sqrt(-407))/2 and the result would be different.> Let's
recall what I posted.> |It appears that Q(sqrt(-167)) has a
class group of order 11.> |Let r stand for (1+sqrt(-167))/2
(which is an algebraic integer,> |in spite of the 2 in the
denominator).> |> |r^11 = (-592764018-86559857*r)> | =
(44555-222*r) (-12882-2017*r)> |> |and> |> |7^11 =
(44555-222*r)(44555+222*r).> As an intermediate step, I found
the GCD of ((1+sqrt(-167))/2)^11> and 7^11, which is an 11-th
power of the GCD of (1+sqrt(-167))/2> and 7.> Harris to check
the result using a calculator or something like that.> |Also
(-12882-2017*r)(-12882-2017*(1-r)) = 2^11 * 3^11, which is>
|relatively prime to 7^11.> |> |The factor 44555-222*r = 44444
- 111 sqrt(-167) and its conjugate> |44444 + 111 sqrt(-167) are
roots of x^2-88888*x+7^11=0. One GCD of> |r with 7 is then
(44444-111 sqrt(-167))^(1/11), which is a root of> |x^22 -
88888 * x^11 + 7^11 = 0.Well I figure you're not lying here
Keith, so I'll go with youranalysis.So I was wrong about being
able to find a factor in the ring ofalgebraic integers.
Interesting.===
===
Subject:
: branch of log zshowing, what's inside
the glass-house, you were talking about.It really helps !Keep
the glasses clean and un-veiled.Keep expressing, what You
really think -there are not too many likeYou.Hero===
===
Subject:
:
Re: Dik Winter's claims revisited, dependency issue|Also
(-12882-2017*r)(-12882-2017*(1-r)) = 2^11 * 3^11, which
is|relatively prime to 7^11.||The factor 44555-222*r = 44444 -
111 sqrt(-167) and its conjugate|44444 + 111 sqrt(-167) are
roots of x^2-88888*x+7^11=0. One GCD of|r with 7 is then
(44444-111 sqrt(-167))^(1/11), which is a root of|x^22 - 88888
* x^11 + 7^11 = 0.> OOPS! I'm re-thinking my previous post.
I'll consider Keith Ramsay's> points here some more.> For all
I know I'm wrong, but then again, maybe I'm still just
missing> something.Both!===
===
Subject:
: Re: How big can a manifold
be?well its emmbeddable in R^3 its obvious of cardinality c,
no?===
===
Subject:
: Re: A newbie's question -- about real number>
But I *can* find an uncountable number of non-terminating>
representations of 89/100>You can't count to 1?Of course not!
Given r1<>r2 it's clear that 1, like 89/100, is
irrational(provided the rationals are not
uncountable).===
===
Subject:
: Re: A newbie's question -- about real
numberHence the rationals are uncountable! > Nice try, but no
cigar. The set of decimal sequences that terminate in all >
0's or all 9's is countable.I think you missed the point,
which I took to be that a false staement implies any
statement.===
===
Subject:
: Re: Simple numbers> Well what is that
algorithm?Checkhttp://www.fpx.de/fp/Software/Sieve.htmlhttp://
www.math.utah.edu/~alfeld/Eratosthenes.htmlhttp://
www.math.utah.edu/history/eratosthenes.htmlhttp://www.utm.edu/
research/primes/prove/prove2_1.htmlhttp://math.nmsu.edu/~
pmorandi/CourseMaterials/FindingPrimes.htmlhttp://
primes.utm.edu/glossary/page.php?sort=SieveOfEratosthenes===

===
Subject:
: Re: errors in an argument by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
i1ILngY04088;> My favorite is birds : How can one evolve into
flying ? So many things> have to happen in order to create
functional wings it seems impossible any> awkward animal
developing in this direction would survive it's cost, or> any
failing flying attempts. The evolutionist arguments sound
totally> absurd : they ran fast or jumped from tree to tree
and got an advantage> slowly developing wings. Sounds like
absolute voodoo to me.>That depends on the size of the animal.
If the animal is small enough,>the terminal velocity can result
in a non-lethal impact with the>ground. (I believe the
threshold is around the size of a mouse or a>cat.) Incremental
levels of control over the landing point then have an>obvious
benefit. >-- >Daniel W. Johnson>panoptes@iquest.net><a
href=http://members.iquest.net/~panoptes/>http://
members.iquest.net/~panoptes/</a>039 53 36 N / 086 11 55
WThat's highly unlikely. As far as I know the first flying
creatures they talk about are dinosaurs, no less. And quite
big ones as well. In any case, a functional wing on a mouse,
that allows even such a manuver is still something very hard
to produce simply by mutation. This is a very hard one to
crack for the theory of evolution. I still didn't hear a good
solution for it.===
===
Subject:
: Re: A newbie's question -- about
real number>Hence the rationals are uncountable! >Nice try,
but no cigar. The set of decimal sequences that terminate in
all >0's or all 9's is countable.> No cigar? Bummer. But I
*can* find an uncountable number of non-terminating>
representations of 89/100 Not in any standard form of basal
notation.> and figured that since r1<>r2 the conclusion
followed.If r1 = 0.89 and r2 = 0.88999... = 0.88 +
.001*sum[i=0..oo, 9/10^-i],then r1 = r2 If that is not true
then clearly 89/100 is irrational! Or you Are.===
===
Subject:
: Re:
What is the Origin of Space and Time?>What is the origin of
space and time?> Where and when could space and time have
originated, if there was no> space and time before they
originated?What would where and when mean before they came
into existence?-- Darryl L. Pierce
<mcpierce@myrealbox.com>Visit the Infobahn Offramp -
<http://mypage.org/mcpierce>What do you care what other people
think, Mr. Feynman?===
===
Subject:
: Re: there is no such thing as
infinity> And can be written in exponentially more obfuscated
a manner than even C> or assembler... :)> Assembler? Bah!! Raw
machine language burnt into a ROM is the only way> to find the
largest number. Ah, I remember when I first started
programming. All I had was a really hotneedle and had to write
my code directly to the SIMMs. It was a major leapforward when
I upgraded and could just type copy con file.exe...> In any
case, I have the world's largest> number written on my office
blackboard, so you all might as well give up.Actually, *I*
have the largest number: it's written in 0.1pt type on
theoutside of my 2-story house, covering all four walls, the
deck, the porchrecalculate! -- Darryl L. Pierce
<mcpierce@myrealbox.com>Visit the Infobahn Offramp -
<http://mypage.org/mcpierce>What do you care what other people
think, Mr. Feynman?===
===
Subject:
: Re: there is no such thing as
infinity> Assembler? Bah!! Raw machine language burnt into a
ROM is the only way> to find the largest number. In any case,
I have the world's largest> number written on my office
blackboard, so you all might as well give up.> Add one to it
and see what the sum is.How about just raising it to something
greater than 1? :)-- Darryl L. Pierce
<mcpierce@myrealbox.com>Visit the Infobahn Offramp -
<http://mypage.org/mcpierce>What do you care what other people
think, Mr. Feynman?===
===
Subject:
: Re: No Set Contains Every
Computable NaturalRussell Easterly says...> There is no TM
that can decide if the string has an> infinite string of
1's.> Has anyone claimed otherwise?>People have claimed the
set of all natural numbers>is a recusive set.The usual notion
of recursive set is that a set S isrecursive as a *subset* of
the naturals if there is aTuring machine T such that for any
*natural* number n(in unary notation, to be specific)> The set
of all natural numbers is not a proper subset of> the set of
all natural numbers. T(n) halts and outputs 1 <-> n is an
element of S T(n) halts and outputs 0 <-> n is not an element
of SIf x is some input that is not a representation of a
naturalnumber, then there is no constraint on what T(n) does.>
According to this definition, the set of all natural numbers>
is recursive because there exists a TM that doesn't even> read
the input and always outputs a 1.I think Darryl's definition
has confused you. The use of integers indefining languages is
not as popular as it once was. Modern books are morelikely to
define a language as a subset of E*, where E is a finite set
ofsymbols. In Darryl's definition, every natural n is
equivalent to a singleunique string s from E*. A TM outputting
1 after reading n = A TM acceptings. A TM outputting 0 after
reading n = A TM rejecting s.In other words, we may have an
alphbabet {0,1}, where (letting e = the emptystring):n -> s
table for N -> {0,1}*------------------------------1 -> e ; 2
-> 0 ; 3 -> 1 ; 4 -> 00 ; 5 -> 01 ; 6 -> 10 ; 7 -> 11 ; 8 ->
000; 9 -> 001 ;10->010 ; 11 -> 011 ; 12 -> 100 ; 13 -> 101 ;
14 -> 110 ; 15 -> 111 ;.......etc....So Darryl's version of
{1^n | n in N}, would contain {3, 7, 15, ....}. Itwould not
neccessarily be every single natural number.l8r, Mike N.
Christoff===
===
Subject:
: Re: Teaching philosophyI promise to
cheerfully accept students who don't have theprerequisites and
can't add fractions without a calculator or solvelinear
equations. I will have them work in groups so,
hopefully,someone in each group can do the work. I will be
their best friend sothey don't think of me as a teacher. We
will have nice pleasantclasses where we all discover things
together so nobody feels left outor has their selfsteem
bruised. Before each exam I will give out apractice exam
****very**** similar to the real exam so they will knowwhat to
expect. I will be very generous with part credit for thosethat
still don't get it, even giving re-takes if necessary.
I**guarantee** a very high success rate in my classes. There
will be somany A's and B's that I will get awesome student
evaluations. Studentswill flock to my classes because I am
such a great teacher. The Deanwill notice the improvement in
my department's success rate and willprobably increase it's
funding. Over time, as my methods catch on,even the University
administration and maybe even the Regents willhold our
department up as a shining example of what is being done
toimprove higher education.> Hah, pretty good!thereby showing
he stands with the students & not the academic snobs.-- Gerry
Myerson (gerry@maths.mq.edi.ai) (i -> u for email)===
===
Subject:
:
Re: No Set Contains Every Computable NaturalRussell Easterly
says...>Using Daryl McCullough's definition, I can claim the
set of>all natural numbers is (1,2,3). Since the string 1111
does>not represent a natural number in my system, it is too
large>to be the input to a Turing machine.How does my
definition imply that the set of all naturalnumbers is
{1,2,3}?If you want to be more general than the usual
treatment,you can do the following: Let A be some finite
alphabet (containing 0 and 1, at least), and let A^omega be
the set of infinite sequences of elements of A. Every such
infinite sequence is a possible state for the infinite tape of
a Turing machine. Now, let S be any subset of A^omega, and let
R be any subset of S. Then we can say that R is recursive
relative to S (my own terminology, not any kind of standard)
if there is a Turing machine T such that for any tape state x:
1. If x is an element of R, then T(x) halts and outputs 1. 2.
If x is an element of S, but not an element of R, then T(x)
halts and outputs 0. 3. If x is not an element of S, then
there is no restriction on what T(x) does. We can also define
a notion of being recursively recognizable relative to S: R is
recursively recognizable relative to S if R is a subset of S
and if there is a Turing machine T such that for any tape
state x: 1. If x is an element of R, then T(x) halts and
outputs 1. 2. If x is an element of S, but not an element of
R, then T(x) either halts and outputs 0, or never halts. 3.
Again, if x is not an element of S, then there is no
restriction on what T(x) does.Of particular interest is the
case when S = the set of unaryrepresentations of natural
numbers (that is, each tape statein S consists of a finite
number of consecutive 1s followedby an infinite number of
consecutive 0s). In that case, ifR is a subset of S, then R is
recursive relative to S if andonly if R is the set of unary
representations of a recursiveset of natural numbers. R is
recursively recognizable relativeto S if and only if R is the
set of unary representation ofan r.e. set of natural
numbers.As you point out, the set of unary representations of
naturalsis not recursive (or recursively recognizable)
relative to theset A^omega of all tape states.===
===
Subject:
: Re:
Mean Value Theorem>Two places I've seen the equivalent
statement: If f is continuously>differentiable on [0,1], then
|f(x)-f(y)|<=sup|f'(u)||x-y| where the>sup is extended over
0<=u<=1.>I can't see why we need f' to be continuous. It seems
by the MVT we>have> f(x)-f(y)=f'(c)(x-y) for some c in
[0,1],>Right.>so that> |f(x)-f(y)| = |f'(c)||x-y|, for some
c in [0,1],> <= sup_u|f'(u)||x-y|.>But without continuity (or
something extra beyond differentiability)>how can you conclude
that the derivative f' is bounded on [0, 1]? You>have to have
boundedness in order to justify your last step...>If f' is not
bounded, then we can assign the value Infinity to
sup_u|f'(u)|>and then the relation still holds.>So it seems
that the differentiability condition is there purely
for>technical reasons (read: rigor), to avoid having to deal
with Infinity>because it does not belong to R.> [Continuing
the rambling]> Well we can always, compactify R by adjoining
-infty and infty, so> that should not be a problem. But what
if x = y? We get> 0 <= infty 0> What to make of that?>
[Stopped the rambling]> With my > G. RodriguesThat's a good
point, but perhaps we can slip in an epsilon ad
hoc.===
===
Subject:
: Re: Teaching philosophy>thereby showing he
stands with the students & not the academic snobs.Yeah, I
figured someone would catch that. Is & a word?
:-)--Lynn===
===
Subject:
: Request for ideas on tricky little
problemConsider a bus route with k stops. When the bus starts
its journey, ppersons are on the bus. Person j wants to get
off at stop s_j<=k. Alsos_j>=1. However the bus cant stop at
more than z < k stops. So a personwho wants to get off at a
stop s_j which is not among the z stops has to getoff at the
nearest one and then walk. Let f(x_j) = (x_j)^3 be the
distanceperson j has to walk to get to his destination, where
x is the number ofstops he has to walk to get to his
destination. Let T be the sum f(x_1) +f(x_2) + ... + f(x_p)The
problem is now to construct a method(algorithm) where the bus
stops atstops, so that the T is minimized. Any ideas on
adecent strategy here ?===
===
Subject:
: Re: A newbie's question --
about real number>lisong@iastate.edu asks this about real
numbers:>Two real numbers r1 = 0.89, r2 = 0.889999999.....
(with infinite 9s), >Theoretically NO.Uh, no, the answer is
YES.>In real life measurement or result calculations in which
only two significant>figures are reasonable, then r1=r2. What
does that have to do with _real numbers_?(Heh, I just realized
that you might be fooled by the wordreal. In fact the real
numbers are not real in a usualsense of the word, they're
mathematical abstractions.)>The context must be appropriate.
>G C************************===
===
Subject:
: Re: Help needed -
primes program| Well how about this:|
--------------------------------------------------------------
--------------| Program PB;|| Var| x,y:word;| f:word;|
i,n:word;| ok:boolean;|| Begin| Write(Enter X and Y -> ');|
Readln(x,y);|| For i:=x to y do| Begin| f:=3;| ok:=true;| If
(i=1) or (i mod 2 = 0) then| Begin| f:=i+1;| ok:=false;|
End;|| While f<=sqrt(i) do| Begin| If i mod f = 0 then| Begin|
f:=i;| ok:=false;| End| else| f:=f+2;| End;|| If ok=true then
Write(i,' is a prime. ');| End;|| Readln;| End.|
--------------------------------------------------------------
--------------| It ignores all number with form k*2 ,
k{2,3,4,...} and it goes till sqrt(i)| insted of i.|| Is this
faste or slower than sieve of eEatosthenes? Why?The above
program doesn't show two (2) as a prime.
__________________Gerard S.===
===
Subject:
: Re: 'erf' function in
C> If you define> R(x) = cPhi(x)/phi(x)> then the function R
is a well behaved function that has> a rapidly convergent
Taylor expansion, an expansion> with terms easily determined
by a two-step recursion.> (Try to derive the recursion
yourself:> r[k+1](x)=x*r[k](x)+k*r[k-1](x),> where r[k](x) is
the kth derivative of R(x).)> ...> double Phi(double x)> {long
double s,t=0,a=1.2533141373155L,z=0,b=-1,pwr=1;> int i;>
s=a+b*x;> for(i=2;s!=t;i+=2)> { a=(a+z*b)/i;>
b=(b+z*a)/(i+1);> pwr=pwr*x*x;> t=s;> s+=pwr*(a+x*b);> }>
return 1.-s*exp(-.5*x*x-.91893853320467274178L);> }It is not
clear what role z is supposed to play here, since it
isinitialized to 0 and never modified. Removing z, and the
termsinvolving z, corresponds to your definition of
r[k](x):double Phi(double x){long double
s,t=0,a=1.2533141373155L,b=-1,pwr=1; int i; s=a+b*x;
for(i=2;s!=t;i+=2) { a/=i; b/=(i+1); pwr=pwr*x*x; t=s;
s+=pwr*(a+x*b); } return
1.-s*exp(-.5*x*x-.91893853320467274178L);}A possible
improvement is to note that the even part of
R(x),(R(x)+R(-x))/2, is equal to 1/phi(x), thus only the odd
part of R(x),(R(x)-R(-x))/2, needs to be computed:double
Phi(double x){long double s,t=0,b=1,pwr=x; int i; s=x;
for(i=2;s!=t;i+=2) { b/=(i+1); pwr=pwr*x*x; t=s; s+=pwr*b; }
return .5+s*exp(-.5*x*x-.91893853320467274178L);}I've been
using a similar algorithm, using R(x) =
1/(Phi(x)*Phi(-x)),which also has a nice Taylor series
expansion with a couple of otherdesirable properties: it is an
even function, thus only the terms inx^2i need to be computed;
and all the coefficients are positive, thuseliminating the
loss of precision resulting from subtracting two largenumbers
to get a small answer. This leads to (2*Phi(x)-1)^2 = 1
-4/R(x), thus Phi(x) = 1/2 +/- sqrt(1/4-1/R(x)), where the
sign of thesqrt should be the same as the sign of x. This has
the advantage thatwhen |x| is large, R(x) is also large, thus
1/R(x) is quite close to 0and the result is fairly close to
the correct value in absolute terms.===
===
Subject:
: Klingon,
anti-universals etc. (Re: the anticlassicalist }{ i:
linguistic negation)> Okrand says he violated a few human
language universals in inventing> Klingon, since of course it
wasn't meant to be a human language. I've got the Klingon
dictionary (which is actually a handbook),I've got the tapes,
I've got the later Klingon Berlitz Guide(not its true title,
it's the Klingon dictionary again,adapted for galactic
hitchhikers, can't remember its realtitle), I've been through
the lot, I haven't come across anything even remotely weird.
If you want something reallyweird, try Lojban instead.(No
apologies for cross-posting to alt.phil, sci.m, sci.l,sci.p,
someone out there started this , not me)===
===
Subject:
: Re:
Dik Winter's claims revisited, dependency issue> So I was
wrong about being able to find a factor in the ring of>
algebraic integers. Interesting.Now go back and count how many
times this was proved to you, and count how many times you
accusedother posters of being liars, co-conspirators, etc. for
posting the truth. Then bump the OOPS!counter.How many times do
you have to be smacked with the 2x4-of-truth before it leaves a
dent?> James Often in error, but never in doubt. HarrisWacky,
isn't it? But, hey, it's just math. Yup, yup, yup.Mixing
medications can be dangerous. Never take a strong laxative and
a sleeping pill at the
sametime..http://www.crbond.com===
===
Subject:
: Re: the
anticlassicalist }{ i: linguistic negationThe not-house will
be built> Is this to mean the not-house won't be not-built?No,
it's a solemn undertaking to build a trampoline.Some argue no,
not a trampoline, but a bridge,which is not a house either. I
counter thattrolls are known to dwell under bridges
(LordBritish, pers.com.), and therefore a bridge is ahouse
since it houses trolls. So a trampoline it must be.===
===
Subject:
:
Re: the anticlassicalist }{ i: linguistic negationKlingon was
specifically created to be the worst language possible byfolks
who knew linguistics. It was enthusiastically embraced by
themob and it is as good as Korean or Chinese for transferring
content.> Okrand says he violated a few human language
universals in inventing> Klingon, since of course it wasn't
meant to be a human language. Nothing> about worst language
possible. Studies of how Klingon-users actually> speak might
be interesting. Do you know of any? And anyhow what was this>
meant to prove, even if it were all true? Sure as hell
wouldn't prove> whatever you were salivating about in the
previous paragraph anyway.You can find a significant community
of people who speak Klingon atthe Klingon Language Institute.
http://www.kli.org/The archives of the KLI's email discussion
forum might be useful ifyou're serious about studying how
Klingon is spoken. http://www.kli.org/tlhIngan-Hol/As for
worst language possible, that would be INTERCAL. Klingon
isactually a very easy language. Compared to natural
languages, it'sessentially a toy, but it's a very powerful and
interesting toy.===
===
Subject:
: Re: errors in an argument> As far
as I know the first flying creatures they talk about are >
dinosaurs, no less. And quite big ones as well.That's a
comment on your knowledge. Some dinosaurs were small; a
quicksearch turned up some with a mass below 10 kg. And all
the references Ifound talk about small dinosaurs as the
starting point for the evolutionof wings.--
http://members.iquest.net/~panoptes/039 53 36 N / 086 11 55
W===
===
Subject:
: Re: Request for ideas on tricky little problem>
Consider a bus route with k stops. When the bus starts its
journey, p> persons are on the bus. Person j wants to get off
at stop s_j<=k. Also> s_j>=1. However the bus cant stop at
more than z < k stops. So a person> who wants to get off at a
stop s_j which is not among the z stops has to get> off at the
nearest one and then walk. Let f(x_j) = (x_j)^3 be the
distance> person j has to walk to get to his destination,
where x is the number of> stops he has to walk to get to his
destination. Let T be the sum f(x_1) +> f(x_2) + ... + f(x_p)>
The problem is now to construct a method(algorithm) where the
bus stops at> stops, so that the T is minimized. Any ideas on
a> decent strategy here ?If a choice of bus stops is optimal,
then there can be no improvement, and especially no trivial
improvement. So if you have an optimal choice of bus stops,
and I take a look at lets say three consecutive stops, then I
won't be able to demonstrate an improvement based on these
stops. This can be used as the basis of an algorithm: For i =
3, 4, ..., k, find all sequences of i consecutive stops that
don't allow a trivial improvement. Start with i = 3. Lets say
I choose the first stop at x, the last at z, and the middle
one would be at y for some x < y < z. Try all possible choices
of y, calculate the cost for all people who want to leave at
one of the stops x to z for each choice of y, and you will
find all choices for y that cannot be trivially improved. Do
this for all x and z, and you have all seqences of three stops
that cannot be trivially improved.Continue with i = 4, so you
look at all sequences of stops x, y, z, u. Pick x and u. Then
choose x+2 <= z <= u-1. For each z, choose those values of y
such that (x, y, z) cannot be improved and (y, z, u) cannot be
improved. For each pair (y, z) calculate the total cost of all
people exiting at stops x to u. Find all pairs (y, z) that
minimise the cost. Do this for all x and u, and you have all
sequences of four stops that cannot be improved trivially.
Continue with i = 5, 6 etc. until i = number of stops that the
bus can make. Only difference when i = number of stops is that
you calculate the total cost. I haven't tried this, so I don't
have any idea how the amount of data will grow, but I would
hope that it isn't too bad.===
===
Subject:
: Re: Teaching
philosophy>Yeah, I figured someone would catch that. Is & a
word? &ante Melancholico To &rea (names have been changed to
protect the guilty) & after all is said & done My life has
been a lonely one, & when I sometimes think of you I underst&
your point of view. & often whilst I contemplate, The
time-worn s&s accumulate ...[The rest is at
www.herring.pwp.blueyonder.co.uk/poetry/andante.htm
]===
===
Subject:
: Re: Got a speeding ticket and need to fight
backThe Lord of Chaos (Suresh Deva)A Prince, is supposed to be
intelligent. duh!!!> I thought that when Euler said There is no
prince way to mathematics he> was politely meaning the
opposite? Or by Prince with capital P you do> refer to some
unfamous singer. That one looks intelligent for sure!Hey! no
wonder why most noble prize winners in physics areridiculously
good looking and so are many famous
mathematicians!!!===
===
Subject:
: Re: Got a speeding ticket and
need to fight back> Good one, chief. I hear the WB network is
hiring comedy writers.WB is also looking for lawyers who can
play sidewalk Santas for theirchristmas in july specials. Btw,
WB is also looking for statistcianswho can prove that christmas
is supposed to start in july.-suresh===
===
Subject:
: JSH: Making it
personalI'm concerned about all the histrionics that has
darkened even my ownposts as it's all too personal at this
point.Right now I *despise* posters like Dik Winter, Nora
Baron, and DavidUllrich, which necessarily involves some fear:
they're potent atfighting for their positions.They also just so
happen to ultimately be wrong.Dik Winter keeps up webpages on
some of my OLD failed attempts atproving Fermat's Last
Theorem.That's just bad behavior.Trotting out someone else's
mistakes made years ago, and pushing themon your own personal
webpages is not something that's defensible.It's clearly
making a statement.Dik Winter invites my contempt, and
unfortunately, I have to fear theeffectiveness of such
tactics.It's sad really.I'm going to see about dropping that
fear as I think it gets in theway of research, but right now I
do despise these people becausethey're so contemptible, and so
successful in screwing up thedialogue.===
===
Subject:
: Re: Bound of
a sum> I don't know if if matters to the original poster, but I
think I can> improve the constant to log(4).That's the bound I
got earlier. See ===
===
Subject:
: Klingon misconceptions (Re: the
anticlassicalist }{ i: linguistic negation)> Klingon was
specifically created to be the worst language possible by>
folks who knew linguistics.Nope. It was created to fit the
model of the archetypalKlingons when they were unsullied
baddies (those you'dzap with phasers and photon torpedoes on
your university'sIBM-370). Klingon was designed to be the
worst languageonly in:(1) acoustic phonetics--it had to sound
foul(2) articulatory phonetics--handbooks of Klingon advise
that spraying the audience with spittle is the right way to
pronounce Klingon aspirated stops. Pity Ockrand didn't know
about the interlabial trills of Malekula!Apart from that,
Okrand had a bit of harmless fun picking a word order that few
languages follow,and making up a set of fused pronominal verbal
prefixes. But Greenlandic does that on such agrand scale that,
compared to it, Klingon is asEsperanto is to Sanskrit. Plus,
without even trying, Okrand came up with phonotactics that
madeautosegmentation dead easy (good for
natural-languageprocessing, that). The authors of Lojban tried
hard,and made a mess of it. To sum it up, Klingon is an ideal
artificial language.It is much easier than... you kn*w (blush,
whisper),whilst being just weird enough so that no-one enjoysa
great advantage over anyone else when learning it.===
===
Subject:
:
Re: 3 Squares Covering 1 Circle> For b, what does a C3v
(invariant under 120 deg rotation) > symmetric arrangement
yield? (Place the outermost edge of> the squares tangent to
the circle.)> Here's a similar situation that does better
than the 80.4%> with part (a). Use this 120 deg symmetry, but
place the diagonals> of the square along the radii. The
optimum placement of the> squares (they can be shifted along
the radii) covers 83.0% of> the circle's area. This is the
best that I've found, but I don't> know if it's the best
possible.> To summarize the results so far, the best result
posted for case (a)> (no overlap) is> (a) pi/6 - 1/2 +
sqrt(3)/4 + sqrt(55)/8 + Arccos(3/8) => 2.57003584 =
81.8067816% of the circle,I have an improvement for case (b)
(over Jake's 83.0% configuration).It again has C3 symmetry,
but not D3 symmetry (as Jake's and Hauke'sconfigurations
do).Let one corner of a square be at (-r,0), and the opposite
corner beat (-r + sqrt(2) cos(phi), sqrt(2) sin(phi)) -- i.e.,
the diagonalconnecting (-r,0) to the opposite corner is
oriented at an angle phi.Let the other two squares be the
image of this one under rotationsby 2 pi/3 and 4 pi/3,
respectively. (Note that for phi = 0, thisreduces to Jake's
configuration.)The optimal case has r = rc and phi = phic,
where rc = 0.192008425, and phic = 0.678933044.The area Ac
achieved by this configuration is Ac = 2.61034477 =
83.0898548% of the circle.The value 2.60808705 for Jake's
configuration is a local maximum fora constrained 1-parameter
problem, and the value above is a localmaximum for a
constrained 2-parameter problem. But of greater interestis
whether the values are local maxima for the unconstrained
problem,which has 9 parameters (3 for each square).I looked
into this. Both constrained maxima are indeed critical
pointsfor the unconstrained problem (the gradient of the area
function is 0for each). To assess whether the critical points
are local maxima, weform the Hessian matrix and look at its
eigenvalues. One eigenvaluemust be 0 (corresponding to the
eigenvector that rotates the entiresystem). For Jake's
configuration, there are 5 negative, and 3 positivepositive
eigenvalues, so it is a saddle rather than a local maximum.For
the new configuration I've given, all 8 non-zero eigenvalues of
theHessian are negative. The configuration is indeed a local
maximum ofthe unconstrained problem, and hence a viable
candidate for the titleof global maximum.Note: if you graph
the configuration it may look like some of the cornersof the
squares lie on the circle, but they do not. They are at a
distance0.99803 from the center of the circle.-Jim
Ferry===
===
Subject:
: Re: help with solutions for three questions
from the past contestthanks for all the help. I did get no.2
and no. 3 fine. but no luck on this one.any help will be
appreciated.1. Al and Bob are at opposite ends of a diameter
of a silo in theshape of a tall right circular cylinder with
radius 150 ft. al is duewest of Bob. Al begins walking along
the edge of the silo at 6 ft. persecond at the same moment
that Bob begins to walk due east at the samespeed. The value
closest to the time in seconds when Al first can seeBob is
what? answer: 48===
===
Subject:
: Re: Help needed - primes program|
Well how about this:|
--------------------------------------------------------------
--------------| Program PB;|| Var| x,y:word;| f:word;|
i,n:word;| ok:boolean;|| Begin| Write(Enter X and Y -> ');|
Readln(x,y);|| For i:=x to y do| Begin| f:=3;| ok:=true;| If
(i=1) or (i mod 2 = 0) then| Begin| f:=i+1;| ok:=false;|
End;|| While f<=sqrt(i) do| Begin| If i mod f = 0 then| Begin|
f:=i;| ok:=false;| End| else| f:=f+2;| End;|| If ok=true then
Write(i,' is a prime. ');| End;|| Readln;| End.|
--------------------------------------------------------------
--------------| It ignores all number with form k*2 ,
k{2,3,4,...} and it goes till sqrt(i)| insted of i.|| Is this
faste or slower than sieve of eEatosthenes? Why?No, it's
slower because you are dividing the numbers by all odd
numbers, thebest you could do (here) is just divide the
numbers by all odd primes(handling the even numbers as a
special case).(also rewritten in REXX) and sped it up by a
factor of three:(orginal algorithm in
REXX):________________________________________________________
_____________________/**/ arg x y .; if x=='' then x=1; if
y=='' then y=12345; tell=y<0; y=abs(y)p=0; call time 'R' do
j=x to y f=3; ok=1 if (j=1) | (j//2==0) then do; f=j+1; ok=0;
end isq=$isq(j) do while f<=isq If j//f==0 then do; f=j; ok=0;
end else f=f+2 end if ok then p=p+1 if tell then say j 'is
prime.' endt=format(time('E'),,2)say p 'primes found between'
x 'and' y 'and took' t
'seconds'exit/*-----------------interger squart root
subroutine-----------*/$isq: procedure; parse arg x;
x=trunc(x); r=0; q=1 do while q<=x; q=q*4; end do while q>1;
q=q%4; h=r+q; r=r%2; _=x-h if _>=0 then do; x=_; r=r+q; end
endreturn
r_____________________________________________________________
_________________squart root, which can be defines as
floor(sqrt(x))Note that REXX (the language I used) has no SQRT
function anyway, so thiswould make it faster than if I used a
regular SQRT function.Note that the orginal version has an
error in it that doesn't show the (only)even prime (that is,
2).Here is the faster version (and correct) version of your
algorithm:____________________________________________________
___________________________/**/ arg x y .; if x=='' then x=1;
if y=='' then y=12345tell=y<0 /*show primes if y is neg.
*/y=abs(y) /*use postive value of Y. */p=0 /*number of primes
found. */ox=x /*save original x specified.*/call time 'R'if
2>=x then do /*special case of even prime*/ p=p+1 /*add 1 to
the num of primes*/ if tell then say '2 is prime.'
endx=max(x,3) /*start with 3 as a minimum.*/x=x+(x//2==0) /*if
x is even, add 1 to it.*/ do j=x to y by 2 /*now, step trough
odd nums.*/ do f=3 to $isq(j) by 2 /*divide j by all odd
numbs.*/ if j//f==0 then iterate j /*not prime, try another
num*/ end p=p+1 /*add 1 to the num of primes*/ if tell then
say j 'is prime.' endt=format(time('E'),,2)say p 'primes found
between' ox and y 'and took' t
seconds.exit/*-----------------interger squart root
subroutine-----------*/$isq: procedure; parse arg x;
x=trunc(x); r=0; q=1 do while q<=x; q=q*4; end do while q>1;
q=q%4; h=r+q; r=r%2; _=x-h if _>=0 then do; x=_; r=r+q; end
endreturn
r_____________________________________________________________
_________________The next step in the optimization of the
program is to only divide by theodd primes found so far (up to
the ISQRT of J), instead of all the odd
numbers.______________________________________________________
_________________Gerard S.===
===
Subject:
: Re: the anticlassicalist
}{ i: linguistic negation> As for worst language possible, that
would be INTERCAL.I found
this:http://www.antipope.org/charlie/journo/intercal.html===

===
Subject:
: Re: Making it personal> I'm going to see about
dropping that fear as I think it gets in the> way of research,
but right now I do despise these people because> they're so
contemptible, and so successful in screwing up the> dialogue.>
James HarrisMr. Harris,Given your eight years of human-sewage
on sci.math, can you please summarizeyour research results for
us?Can you please site the number of papers accepted by
journals and actuallysite those papers?Do you realize that you
are wasting your life?You have called mathematicians names,
scolded people who know more than youdo, lied, performed chest
beating feats the likes of which no one has everseen and where
has all of that led you?Are we to feel pity for you?Are we to
feel sorry for you?Do you realize that you are in need of
professional help?You have brought upon yourself any and all
criticisms only by the path youhave chosen.Don't cry now,
eh?===
===
Subject:
: Re: Making it personal> I'm concerned about all
the histrionics that has darkened even my own> posts as it's
all too personal at this point.> Right now I *despise* posters
like Dik Winter, Nora Baron, and David> Ullrich, which
necessarily involves some fear: they're potent at> fighting
for their positions.> They also just so happen to ultimately
be wrong.> Dik Winter keeps up webpages on some of my OLD
failed attempts at> proving Fermat's Last Theorem.> That's
just bad behavior.> Trotting out someone else's mistakes made
years ago, and pushing them> on your own personal webpages is
not something that's defensible.> It's clearly making a
statement.> Dik Winter invites my contempt, and unfortunately,
I have to fear the> effectiveness of such tactics.> It's sad
really.> I'm going to see about dropping that fear as I think
it gets in the> way of research, but right now I do despise
these people because> they're so contemptible, and so
successful in screwing up the> dialogue.> James HarrisIf you
make it known you're pissed about Dik's webpage, he's probably
notgoing to change it. Besides, you exhibit the same bad
behavior when you callus names and such.David Moran===
===
Subject:
:
Re: Teaching philosophy>I need to write a description of my
teaching philosophy. In order to>do so accurately, I think I
need to write at greater length and in>greater detail on this
topic than any hiring committee will want to read.>Moreover, I
need to present my views accurately but, somehow, in such>a
manner as not to vitiate the consideration of my
application.>That being the case, can someone please tell me
what my teaching>philosophy is?>Ignorantly,>Allan Adler> A
more serious answer:I promise to cheerfully accept students
who don't have the> prerequisites and can't add fractions
without a calculator or solve> linear equations. I will have
them work in groups so, hopefully,> someone in each group can
do the work. I will be their best friend so> they don't think
of me as a teacher. We will have nice pleasant> classes where
we all discover things together so nobody feels left out> or
has their selfsteem bruised. Before each exam I will give out
a> practice exam ****very**** similar to the real exam so they
will know> what to expect. I will be very generous with part
credit for those> that still don't get it, even giving
re-takes if necessary. I> **guarantee** a very high success
rate in my classes. There will be so> many A's and B's that I
will get awesome student evaluations. Students> will flock to
my classes because I am such a great teacher. The Dean> will
notice the improvement in my department's success rate and
will> probably increase it's funding. Over time, as my methods
catch on,> even the University administration and maybe even
the Regents will> hold our department up as a shining example
of what is being done to> improve higher education.> --LynnOh,
sod the it's its confusion, this is the funniest and most
heartwarming thing I've ever read in this group.I had started
a reply based on my experiences there, but thought ittoo much
like sour grapes.===
===
Subject:
: Re: Peer ReviewAnother anecdote:
Simon Singh in his book Fermat's Last Theorem notes what may
havebeen the first instance of peer review injustice. Hippasus
in theschool of Pythagoras realized that irrational numbers
exist. This wasregarded as heresy and he was put to death for
it.===
===
Subject:
: Re: question about periodic function>If you
have two periodic functions f(t) and g(t) with periods a and
b,>then the function f(t) + g(t) is periodic with period c
where c is the>least-common-multiple of a and b. Here the
first period is 1/.018=500/9 nd>the second period is 1/.02=50.
So the least-common-multiple is 500, i.e.>1/.002. That should
be the period of the combined function.>How to prove the
theorem? Any hint is welcome. TIA.Don't read too much into
this. It's a period, not the period, thatis lcm(a,b): there
might be a smaller period. All that's involved hereis the
rather obvious fact that any common multiple of a and b is a
period of both f(t) and g(t), and thus of f(t)+g(t). For
example, f(t) = cos(t/2) + cos(t/3) has period 12 pi andg(t) =
cos(t/5) - cos(t/2) has period 20 pi. So the lcm of 12 pi and20
pi, namely 60 pi, is a period of f(t) + g(t). But f(t) + g(t)=
cos(t/5) + cos(t/3) also has period 30 pi.===
===
Subject:
: Re: How
to choose a matrix P>No D!=E>so P can't be an identity
matrixBut P = I works no matter what D and E are. Maybe you
meanY1 != Y2?>Now I am thinking of this question.>Suppose
>Y1=D - P'EP >Y2=D >where D,E,P : nxn matrix>||P||=1 >How to
choose P, so that >rank Y1 >= rank Y2 >and >||Y1||>= ||Y2|| >
The identity matrix would seem to work...> Maybe you have some
more requirements you're not telling us. Robert Israel
israel@math.ubc.ca Department of Mathematics
http://www.math.ubc.ca/~israel University of British Columbia
Vancouver, BC, Canada V6T 1Z2===
===
Subject:
: Re: JSH: Making it
personal > Right now I *despise* posters like Dik Winter, Nora
Baron, and David > Ullrich, which necessarily involves some
fear: they're potent at > fighting for their positions.Because
we understand the mathematics perhaps? > They also just so
happen to ultimately be wrong.Are we? > Dik Winter keeps up
webpages on some of my OLD failed attempts at > proving
Fermat's Last Theorem. > That's just bad behavior.I have
explicitly said *why* I keep them up. As your current way
todiscuss is just exactly the same as is succinctly shown on
thosepages I see no reason to pull them down. It is your
discussions(or actually lack of discussions) that are shown.
And your way todiscuss it just bad behaviour, I would say. >
Trotting out someone else's mistakes made years ago, and
pushing them > on your own personal webpages is not something
that's defensible.Well, the kind of mistakes you made back
than are similar to the kindof mistakes you make now.
Moreover, the way you respond to peoplepointing out the
mistakes has also remained the same (although currentlyyou are
quite a bit more foul-mouthed). > It's clearly making a
statement.Indeed. You are inept at handling contradictions to
what you think istrue. As has been shown in the last 6 weeks
over a common factorbetween two numbers. I may note that the
*existance* of such a commonfactor was already shown much
earlier. At least October last year. > Dik Winter invites my
contempt, and unfortunately, I have to fear the >
effectiveness of such tactics.Oh, well. Do not fear, I will
not hurt you. > It's sad really.Yup. > I'm going to see about
dropping that fear as I think it gets in the > way of
research, but right now I do despise these people because >
they're so contemptible, and so successful in screwing up the
> dialogue.I thought you were the screwer. You never verify
what is explicitlystated, you only say it is wrong.-- dik t.
winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland,
+31205924131home: bovenover 215, 1025 jn amsterdam, nederland;
http://www.cwi.nl/~dik/===
===
Subject:
: Re: How many ways to put 5
balls into 500 ordered cups?> No fair doing people's homework
for them :-) <<Hey! I'm faculty! I'm just slow :)===
===
Subject:
:
Re: How many ways to put 5 balls into 500 ordered cups?> Stars
and bars proof: .. <<Looks great! Send me an email with your
snail mail and I'll get youon the comp list for the new book,
as promised.===
===
Subject:
: Re: Coprime Grid: Filling Infinite
Quadrant>In this post, I write of a specific and altered case
of the puzzle>mentioned in these previous threads:http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&safe=off&
threadm=b4be2fdf.0211011659.79913415%40posting.google.com&rnum
=3&prev=http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&safe=off&
threadm=b4be2fdf.0401241437.73a12678%40posting.google.com&rnum
=4&prev=http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&safe=off&
threadm=b4be2fdf.0303071814.63262520%40posting.google.com&rnum
=1&prev=>In the puzzle, we try to write, in order, the
positive integers into a>grid, one integer per grid-square,
such that:>Each integer n is adjacent (above /left of /right
of /below)>of the integer (n+1);>And each adjacent pair of
integers (above /left of /right of /below)>are coprime.>And
the goal is to completely fill the grid.>But here I am asking
about filling an infinite grid which is bounded>along 2
perpendicular sides.>ie. the grid is an entire quadrant of the
Cartesian plane, bounded by>the x-axis and the y-axis.>I think
I found a simple procedure which *might* ensure a
successful>filling of the grid with coprime-adjacent
integers.>(Sorry to those on rec.puzzles, but I will give my
procedure below.>You can still post your own algorithm,
however, or confirm that mine>can really work for the entire
grid without problems.)>I illustrate with the first 99
terms:>(figured by hand, so maybe inrror)>99>98>97 96 95>54 55
94 93 92>53 56 57 58 91 90 89>52 51 50 59 60 61 88 87>21 22 49
48 47 62 63 86>20 23 24 25 46 45 64 85>19 18 17 26 27 44 65
84>06 07 16 15 28 43 66 83 82 81 80 79>05 08 09 14 29 42 67 68
69 70 71 78>04 03 10 13 30 41 40 39 38 37 72 77 76>01 02 11 12
31 32 33 34 35 36 73 74 75>Basically, the path swings
clockwise and counterclockwise, running>along the outside of
the already-filled section.>When it gets to either the x-axis
or y-axis, it forms a 'peninsula',>the length of which is the
shortest needed to avoid uncoprime integers>being placed next
to each other in the path-section which runs from>that
peninsula's axis to the other axis.>Now, we do not want a
situation where, following the algorith>precisely,>there is NO
peninsula-length which would avoid uncoprime neighbors.>I am
not certain, but I believe this issue is not a problem.>Fun
perhaps: Show if my algorithm is foolproof...or just foolish.>
Well, what I know about number theory and 50 cents will get me
a cup> of coffee, but it seems to me that the way to attack
this solution> is to show that once the peninsula length
exceeds some number N,> you will always have a noncoprime pair
when it loops its way back.> It seems to me that this ought to
be true. > Of course, that stil wouldn't prove that your
algorithm can't> work, because you might never have to buld a
peninsula out that big,> but it would be a start.>(I know I do
not give the peninsula's length. Also fun perhaps: try
to>determine the shortest length needed for each peninsula,
given that>the algorithm never leads to a problem.)>Leroy
Quet> GeorgeI do not believe you are necessarily right about
the filling-algorithmworking for all peninsula-lengths above
N(k), for the k_th peninsula.For if we lengthen the peninsula
by r, each integer following thepeninsula in the path is
increased by 2r.But, for p = any ODD prime, and integer n in
the path after thelengthened peninsula, n is originally
congruent to m (mod p).And after adding (2r), n becomes
congruent to (m+2r)(mod p),obviously.But if n was originally
next to a multiple of p (so in the originalpath, m was coprime
with p), then we get after adding 2r, n + 2r == m+2r (mod p),
and(m+2r) is divisible by p for an infinite number of r's,
given that pis odd.Any progress by anyone out there, or
anything more of interest toadd?...(Yes, yes, I know: It is
not how long your peninsula is, but how youuse it...)Leroy
Quet===
===
Subject:
: Re: RFI:Wanless' Fourth Conjecture/Dirichlet's
Geometric Theorem> _If_ it's true - can one prove it from
Dirichlet's [Arithmetic ie standard]> Theorem?A) Regardless of
its truth or falsity, I am sure that you can prove it from
Dirichlet's Theorem.B) Consider a=234578273658273645827341
b=2348798769847962341> J>The series a^n+b [hcf(a,b)=1]
includes an infinite number of primes [over>all n]===
===
Subject:
:
Re: No Set Contains Every Computable Natural> Russell Easterly
says...>Using Daryl McCullough's definition, I can claim the
set of>all natural numbers is (1,2,3). Since the string 1111
does>not represent a natural number in my system, it is too
large>to be the input to a Turing machine.> How does my
definition imply that the set of all natural> numbers is
{1,2,3}?We have already agreed there is no way a TM can
decideif the number of symbols on a tape is finite(in a finite
number of steps).If I get to decide what is finite, then I can
say anythingwith three or less symbols is finite and anything
withmore than three symbols is infinite.Of course, most people
would use the standarddefinition of N we all know and love.> If
you want to be more general than the usual treatment,> you can
do the following:> Let A be some finite alphabet (containing 0
and 1, at least),> and let A^omega be the set of infinite
sequences of elements of A.> Every such infinite sequence is a
possible state> for the infinite tape of a Turing machine.>
Now, let S be any subset of A^omega, and> let R be any subset
of S. Then we can say> that R is recursive relative to S (my
own> terminology, not any kind of standard) if there is a>
Turing machine T such that for any tape state x:> 1. If x is
an element of R, then T(x) halts> and outputs 1.> 2. If x is
an element of S, but not an element of R,> then T(x) halts and
outputs 0.> 3. If x is not an element of S, then there is no>
restriction on what T(x) does.> We can also define a notion of
being recursively> recognizable relative to S: R is recursively
recognizable> relative to S if R is a subset of S and if> there
is a Turing machine T such that> for any tape state x:> 1. If x
is an element of R, then T(x) halts and outputs 1.> 2. If x is
an element of S, but not an element of R, then> T(x) either
halts and outputs 0, or never halts.> 3. Again, if x is not an
element of S, then there is no> restriction on what T(x) does.>
Of particular interest is the case when S = the set of unary>
representations of natural numbers (that is, each tape state>
in S consists of a finite number of consecutive 1s followed>
by an infinite number of consecutive 0s).There is that word,
finite, again.Can I assume that it is never the case there is
a stringwith an infinite number of 1s followed by an
infinitenumber of 0's?This would mean there is exactly one
string in A^omegathat contains an infinite number of 1's with
nopreceding 0's.> In that case, if> R is a subset of S, then R
is recursive relative to S if and> only if R is the set of
unary representations of a recursive> set of natural numbers.
R is recursively recognizable relative> to S if and only if R
is the set of unary representation of> an r.e. set of natural
numbers.> As you point out, the set of unary representations
of naturals> is not recursive (or recursively recognizable)
relative to the> set A^omega of all tape states.I guess I
should be happy you agree with me.Of course, that would be too
simple.I really don't understand why R is not recursively
recognizable.I warned you that I sometimes contradict myself.I
think I can prove that a TM will say that anystring of
consecutive 1s is finite.This proof also shows there is no
such thing as aninfinite string of consecutive 1's, at least
asfar as a TM is concerned.Before I give a complicated proof,
I should givea simple explaination. Assume you want to countto
infinity. You start with 1, 2, 3, etc.Will you ever get to
infinity? Of course not.You can never add 1 to a natural
number and getinfinity. A TM can add, subtract, multiply,
and,maybe divide (division is kind of iffy). You can'tcompute
infinity using these operations. A TMwill never say a string
is infinite because it isincapable of doing so.Now, the
complicatd proof:I define two TMs, A and B.TM A1) write 12)
move rightrepeat steps (1) and (2)We might think this TM will
write an infinite string of 1s,but it doesn't.TM B1) Find a
blank2) Find a second blank3) Backup and write a 1 on the
previous blankrepeat steps (1) through (3)TM B will write a
finite number of 1s followedby a blank. The number of 1s
written by TM A isa multiple of the number written by TM B.1)
write 1 read blank2) move right move right3) write 1 read
blank4) move right move left5) write 1 write 16) move right
move rightrepeatEither both TMs write a finite number of 1sor
both TMs write an infinite number of 1s.Either way, the output
tape will still containan infinite string of trailing
blanks.(Of course, we can replace the blanks with 0).Russell-
2 many 2 count===
===
Subject:
: Re: What is a number?/What is not a
number? <4030da8f$12$fuzhry+tra$mr2ice@news.patriot.net>
<844a1b64.0402161427.29232391@posting.google.com>
<Ht7DED.Crx@cwi.nl>
<844a1b64.0402170009.534eb1f9@posting.google.com>X-Cise:
tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate:
admin@interspeed.co.nzXcrate: tanandtanlawyers.comX-Pose:
george_cox@btinternet.comX-Punge: Micro$oftX-Sanguinate:
themvsguy@email.comX-Terminate: SPA(GIS)X-Tinguish: Mark
Griffith <markgriffith@rocketmail.com>X-Treme: C&C,DWS>We have
three ways to proceed:More than that.>1) revamp the definition
above to remove polynomials,>2) accept polynomials as numbers,
or>3) define a number set as any mathematical set at all.4)
Define a number as any element of an algebraic structure5)
Don't use the term number at all, except as a constituent of
compound terms, e.g., real number.There are, of course, yet
other options.My prefered option is 5.-- Shmuel (Seymour J.)
Metz, SysProg and JOATUnsolicited bulk E-mail will be subject
to legal action. I reservethe right to publicly post or
ridicule any abusive E-mail.Reply to domain Patriot dot net
user shmuel+news to contact me. Donot reply to
spamtrap@library.lspace.org===
===
Subject:
: Re: Graph Theory
Textbook> I'm an undergraduate, and this summer I will be
participating in an> REU for discrete math and combinatorics.
I am looking for a good> graph theory textbook to learn the
basics from. I have seen the books> that Dover (which of
course are very cheap) has to offer on the> subject, but
unfortunately they seem too elementary. Is the Springer> GTM
Graph Theory by Reinhard Diestel any good? Any suggestions
are> highly welcomed.> JackI learned from a book called
Introduction to Graph Theory by Robin Wilson. It's runs about
$40-50 used, so it's not the cheapest book, especially for
paperback. However, I thought that the exercises were very
appropriate. They were some that required modern algebra and a
few that required beginning combinatorics. Some questiosn were
simple verify problmes and there were a few proofs, but
nothing way over the top. Answers to selected exercises were
in the back, including a few of the proof answers. The proofs
that were covered in the text were, for the most part,
standard proofs for the topics. The book also did a good job
defining and explaining terms. It also has a chapter covering
matroids. For undergrads, you can probably skip that chapter
and not lose anything. However, if you do cover that chapter,
I suggest looking at another book for a more complete
coverage.Oh, the class I took was an upper-level undergrad,
lower-level graduate course. - Tim-- Timothy M. BrauchGraduate
StudentDepartment of MathematicsWake Forest
University===
===
Subject:
: re:intercept length when a random line
intersect with ellipsoidIn order to answer your question, you
need to be more specific aboutrandom for the line. A line in
space can be specified as all pointP satisfying P=X+tV, where
X is a random point in space (needing aprobability
distribution) and V is a random unit vector
(isotropic?distribution), while t ranges over the entire real
line.Once these distributions (for X and V) and the parameters
of the shellare defined, you will have a tractable porblem.
Posted Via Usenet.com Premium Usenet Newsgroup
Services------------------------------------------------------
---- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY
**----------------------------------------------------------
http://www.usenet.com===
===
Subject:
: Integrals without explicit
formulasI am in a differential calculus class. Just about all
the problemsinvolve integrals, but I have been informed that
some of the integralsin these problems have no explicit
formulas. Can you give me any rulesof thumb so I know what
kind of integrals to not bother trying tointegrate?===
===
Subject:
:
Squares Of Certain Areas: puzzleYou have a FIXED sequence of
distinct positive integers, {k}, which isto be determined.For
all k's <= EACH positive integer m, we can take a number n (n
<= m) of (geometric) squares,each of *integer*
side-length,where the j_th square is of the largest squarewith
an integer area that is <= m/(k_j).But {k} is such that, if we
lined the squares up vertically so thattheir
horizontally-running edges coincide and they do not
overlap,the length from the upper side of the top square to
the lower side ofthe bottom square will ALWAYS be m.2
questions: 1) Which integers make up {k_j}?2) Give a formula
for the total area of the (n number of) squares.Example:For m
= 4,We have 1, 2, 3 in {k}, but not 4.So, we have the squares,
as lined up: ------! ! ! ------ k=1, area = 4 <= 4/1! ! !
==----! ! k=2, area = 1 <= 4/2 ==! ! k=3, area = 1 <= 4/3
--Height of the stack is m = 4. Area is 6.Leroy
Quet===
===
Subject:
: If We Replaced Each Prime With -1...Let c(k) =
the sum of the exponents in the prime factorization of k.So, if
we exchanged each prime in the prime factorization of k
with(-1),we would get (-1)^c(k).What I am wondering, however,
iswhat is C(m) =sum{k=1 to m} (-1)^c(k)asymptotical
towards?C(m) also equals:sum{k=1 to m} floor(sqrt(m/k))
*mu(k),where mu() is the Mobius function(or Moebius
function).So, I first thought that sqrt(m) had something to do
with theasymptotics of C(m).But the difference in the sum
before and after taking the floor is toosignificant for me to
claim anything.I might as well write the above equations in
ascii-art mode for thoseexpecting that:;)C(m) = m--- c(k)/
(-1) = ---k=1 m--- _____ | / |/ | V m/k | *mu(k)--- -
-k=1Leroy Quet===
===
Subject:
: Re: Got a speeding ticket and need
to fight backGood one, chief. I hear the WB network is hiring
comedy writers.> WB is also looking for lawyers who can play
sidewalk Santas for their> christmas in july specials. Btw, WB
is also looking for statistcians> who can prove that christmas
is supposed to start in july.It took you six days to come up
with *that*? Pathetic.Doug===
===
Subject:
: Re: Integrals without
explicit formulas> I am in a differential calculus class. Just
about all the problems> involve integrals, but I have been
informed that some of the integrals> in these problems have no
explicit formulas. Can you give me any rules> of thumb so I
know what kind of integrals to not bother trying to>
integrate?First of all, why are you integrating in a
differential calculus class? Thatreally doesn't make sense to
me.Whenever I integrate, I look at what I'm working with.
Polynomials, trigfunctions, log and exponential functions, and
rational functions (though notalways) are usually pretty
straight forward. Rational functions can getmessy, especially
if you're working with partial fractions. Most
functionsinvolving radicals will lead into trigonometric
substitution which will leadto a trigonometric integral. Some
functions (i.e. sin(x^2)) will requirepower series and for
others that you can't integrate, you can use a RiemannSum. I'd
look up the Trapezoidal Rule, as well as breaking the area
intosubintervals and then find the area of each rectangular
subinterval.David Moran===
===
Subject:
: Re: Teaching philosophy> I
need to write a description of my teaching philosophy. In
order to> do so accurately, I think I need to write at greater
length and in> greater detail on this topic than any hiring
committee will want to read.Instead, go for clarity and
brevity.===
===
Subject:
: Re: A newbie's question -- about real
number> Two real numbers r1 = 0.89, r2 = 0.889999999.....
(with infinite 9s), If you add .11 to both sides of the
equation, you get 1 = .9999...thus illustrating the
mathematician's favorite pastime of reducing this problem to
another problem we've seen before . . .===
===
Subject:
: Re: Collatz
Conjecture : Symmetry question.> But there are vastly more
total numbers at a given level than there are> integers, so
although this is true, I'm not sure it leads anywhere.> For
example, I recently found all the integers that are at level
84> (from 1):> 179,441,377> If I included all the potential
non-integer ancestors, I would have had to> sift through 2^84
or> 19,342,813,113,834,066,795,298,816Yes, this is true. But,
taking the fact that within
the19,342,813,113,834,066,795,298,816 values - if looked at as
an orderedlist in magnitude, then the location of integers is
static for allsymmetric trees. Hence, a set of 179,441,377
difference values can beused to calculate all integer values
directly (I am leaving out a biton how to do this but is is
quite easy). However, the continuity ofconnectivity goes out
the window as you start jumping around amoungthe different
symmetric trees. Bummer. It seems there is no end to thewalls
between us and a solution.===
===
Subject:
: Re: JSH: Making it
personal> I'm concerned about all the histrionics that has
darkened even my own> posts as it's all too personal at this
point.You can control your side of it. Put up or shut up.>
Right now I *despise* posters like Dik Winter, Nora Baron, and
David> Ullrich, which necessarily involves some fear: they're
potent at> fighting for their positions.They have been
fighting for the truth. What's *your* position?> They also
just so happen to ultimately be wrong.Nope. You are wrong.>
Dik Winter keeps up webpages on some of my OLD failed attempts
at> proving Fermat's Last Theorem.So what happens if he
replaces them with your NEW failed attempts atproving FLT?
Will you be happy then?> That's just bad behavior.Posting
public records is not bad behavior. Posting false theories
andflawed mathematics, such as you have consistently done, and
launchingpersonal attacks on those who identify your errors
*is* bad behavior.> Trotting out someone else's mistakes made
years ago, and pushing them> on your own personal webpages is
not something that's defensible.Don't post mistakes. In a
public forum anything you post is subject toreview. And what
if he deletes all the mistakes made years ago andreplaces them
with mistakes made recently? Is that better?> It's clearly
making a statement.Duh!> Dik Winter invites my contempt, and
unfortunately, I have to fear the> effectiveness of such
tactics.If the tactic you refer to is making your ridiculous
claims and stupidmistakes easily accessible, it is worthy of
praise. It is *you* who invitecontempt.> It's sad really.Well,
I'd certainly be sad if my ignorance were as easily discovered
asyours is.> I'm going to see about dropping that fear as I
think it gets in the> way of research, but right now I do
despise these people because> they're so contemptible, and so
successful in screwing up the> dialogue.You despise these
people because they have identified errors in yourpompous
claims, and have refused to back down in the face of
yourslanderous gutter language.> James Often in error, but
never in doubt. HarrisThere are two things you must never
attempt to prove: the unprovable --and the
obvious.http://www.crbond.com===
===
Subject:
: Coprime Grid /
Increasingly-Sized StepsHere is a question which combines 2
earlier puzzles of mine into one.As
in:http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&safe=off
&threadm=b4be2fdf.0402151952.6a0480a%40posting.google.com&rnum
=12&prev=we attempt to: Start with an infinite (in every
direction) rectangular grid/lattice.Put 1 in a square of the
grid (or at a vertex of the lattice).Place 2, 3, 4, 5,
....infinity, by some algorithm, into thegrid/lattice so that
each integer (1+m) is exactly m squares (in onlydirections of
either up,down,left,or right) from the square/vertexwith the
integer m, for all m's.And only one, no more/ no fewer,
integer per vertex/square.But...as
in:http://groups.google.com/groups?dq=&hl=en&lr=&ie=UTF-8&
threadm=266426e1.0401280041.5b6ea997%40posting.google.com&prev
=the integers are to be placed so that EVERY pairof adjacent
squares contains two integers which are COPRIME with
eachother...(Adjacent is as defined as immediately next to in
the directions ofeither up, down, left, or right.)A few
things:I am wondering if it can be proved it is possible to
eventually placeall the integers, using the restictions above,
into the infiniteunbounded grid with exactly one integer per
grid-square.My instinct is that there are ways to fill the
infinite gridsuccessfully.And if so, what would the 'most
compact' filling be?By 'most compact', I mean:which placement
of integers would minimize:limit{n -> oo}(area of convex hull
of first n integers' positions)/n? (I could also ask this
question about the less specific infinitevariations of the 2
puzzles from which this one was derived.)Leroy Quet===
===
Subject:
:
Advice on future with MathI'm an upperclass math major that
was/is planning on attempting a masters orphd in math. Here
lately, I've wondered if this is a great idea. My first
question is this: How much should you study (reading the
material,working problems) for, say, an abstract algebra or
advanced calculusjunior/senior level class? I talked to a
graduate student that said he studiedfor a couple of his first
year grad classes at least 4 hours a day on weekdaysand about 8
hours on Saturday and Sunday. Needless to say, I do nothhing
likethat. Should I be doing that or is that just an insane
amount of time to bespending on it? It seems that if you're
good at something, it shouldn'trequire so much work.At times I
have difficulty understanding things that, after finding out
where Iwent wrong, seem pretty simple. I assume others don't
have this problembecause normally they can respond right away
with what some math concept meant or, less often, help me
where I am having a problem understanding part of aproof. Is
it normal to not follow a proof or just a sign that I'm just
weakmathematically (we're again talking junior/senior level
advanced calc/abstractalgebra here)?I have a bad tendency to
be hot/cold when it comes to math. It seems like I'llhave the
highest/near highest score on an exam and then really screw up
on thenext exam, getting something in the bottom half of the
grades. I wish that Icould attribute it to something like when
I first started school: I would dono work until after I screwed
up on the first exam, and then work harder afterthat and do
quite well. Now it seems that I'll do well on the first exam
andthen screw up on the next one. I really don't think that I
will have studiedmuch less for the second one but I guess it
could be a possibility. When I dobad on the other exam, I
begin to wonder if I just got lucky on the problemsthat were
given on the good exam and would have done lousy if a few
differentproblems were picked.Other times, I wonder if I am
just not good at proofs, thus making me about thecrappiest
grad school candidite out there. I sit and stare at the
homeworkproblems and often can't get anywhere. After I see an
answer, it doesn't looktoo bad but it just seems that they're
impossible at times. I had a bad proofsclass that was designed
for just about anyone wanting to take it, so it wasreally easy.
I'm not sure if I can use that excuse at this point though.
Mypre-calculus education was really bad since I was from a
really poor school (Ioften have no clue what certain things
that were assumed as 'knowledge' fromhigh school) but again
this doesn't seem to matter too much in the
proof-basedclasses.Well, that was kind of long-winded. I hope
that I got the point across. Anyadvice would be helpful.
===
===
Subject:
: Book on MATH game theorysense of economics at all,
but a book that deals with actual games, andconcrete
mathematics. I hear there is a good book in the MAA library,
but do not know the name of it.book level should be at the
upper undergraduate / early graduate level in termsof
difficulty but should also be accessible to high school
students with aproblem solving background.All suggestions will
be much appreciated, regards, Toto===
===
Subject:
: Re: Teaching
philosophyteaching philosophy is. Admirable as Lynn's
composition is, Ihave two problems with it:(1) It is Lynn's
point of view (maybe) but not mine.(2) It is actually a
contract of sorts, containing a number of promises, and a
prognosis of sorts for the results of carrying out those
promises. It is not philosophy.Of course, I really have no way
of knowing whether, when a hiring committeeasks for philosophy,
it really means philosophy. That's part of the question.It has
also been pointed out that I have not stated in other terms
whatmy philosophy of teaching is. And one can correctly
conclude thatI am not submitting something here for revision.
But I've been postingto sci.math for a long time, including on
topics related to education,and I was hoping that people who
might have some familiarity with mypoint of view might also be
able to explain the interface between thepositions I've taken
and what the anonymous hiring committees mightbe asking for.It
has also been suggested (privately) that I describe my
teachingexperiences or some of the innovative things I have
tried. But I thinkthat falls under the category of gimmicks or
technique, not philosophyper se.I have no lack of things to
tell a hiring committee. I just have noidea what they are
asking for, nor how much of it.Ignorantly,Allan
Adlerara@zurich.ai.mit.edu************************************
***************************************** ** Disclaimer: I am
a guest and *not* a member of the MIT Artificial **
Intelligence Lab. My actions and comments do not reflect ** in
any way on MIT. Moreover, I am nowhere near the Boston **
metropolitan area. **
**************************************************************
***************===
===
Subject:
: Re: Teaching
philosophy[...snipped...]> Of course, I really have no way of
knowing whether, when a hiring committee> asks for philosophy,
it really means philosophy. That's part of the question.My
impression is that teaching philosophy really means
teachingstatement. Does that help?> It has also been pointed
out that I have not stated in other terms what> my philosophy
of teaching is. And one can correctly conclude that> I am not
submitting something here for revision. But I've been posting>
to sci.math for a long time, including on topics related to
education,> and I was hoping that people who might have some
familiarity with my> point of view might also be able to
explain the interface between the> positions I've taken and
what the anonymous hiring committees might> be asking for.> It
has also been suggested (privately) that I describe my
teaching> experiences or some of the innovative things I have
tried. But I think> that falls under the category of gimmicks
or technique, not philosophy> per se.Well, that's probably
what they want. A good resource
ishttp://www.ams.org/employment/academic-job-search.htmlIn it,
Tom Rishel, a man whose advice you should take seriously,
says,First, because it's easier to write a statement than a
philosophy,let's start with I began teaching in 1993 rather
than Teaching islove and caring. .....Start with concrete
details of your teachingexperience. What courses have you TA'd
or taught and at what levels? How do you conduct a typical
class? Do you have a particularassignment or class that
pleased you and that you wish to describe? Was there a
classroom situation that taught you something meaningfulabout
the way you teach and the way students react to your
methods?Have you checked this resource out before? If not, I'm
rather===
===
Subject:
: Re: I'd like to join this
group!Content-transferncoding: 8bitThat was actually printed
in OMNI magazine Nov 79 , as proof that 2=1 .> ;-)> It was
also printed in the Wall Street Journal to explain how
company> earnings are up. ;-)I am sure GW has used it in his
calculatations.===
===
Subject:
: RudinThis is a pain-in-the-rear
question as it involves looking somethingup in a book. In
Rudin's Principles 3rd. ed. I'm having a problemunderstanding
the inequality near the top of p 305 (section onLebesque
Theory, proof that u*(A)= u(A) for all A in E, E being
thefamily of all elementary subsets of R^p). Specifically I
can't seethe reason forSum(n=1 to N) u(A_n) <= Sum (n=1 to oo)
u(A_n) + epsilon. All help much appreciated.===
===
Subject:
: Re:
Book on MATH game theoryI just happened to be on amazon when
you asked and I tried MAA and gametheory and received:Game
Theory and Strategyby Philip D. Jr
Straffinhttp://www.amazon.com/exec/obidos/tg/detail/-/
0883856379/qid=1077161183/sr=1-2/ref=sr_1_2/102-3578915-
2424159?v=glance&s=booksI'm not sure if that's the difficulty
level you're looking for but it soundslike your description
otherwise.===
===
Subject:
: Re: oil of bitter almonds: Woehler and
LiebigI've made a little progress figuring out what Woehler
and Liebig aretalking about. First of all, they do a
combustion analysis of two samplesof oil of bitter almonds,
namelyI. 0.386 gramme = 1.109 carbonic acid, and 0.200
water.II. 0.341 = 0.982 0.175 which are fairly close to what I
get from their data: I IICarbon ........... 79.438 ..........
79.603Hydrogen ......... 5.756 .......... 5.734Oxygen
......... 14.808 .......... 14.663carbon, hydrogen and oxygen
or, what is the same, the ratio 14:12:2, asWoehler and Liebig
do. How exactly they decided on 14 instead of 7 isn'tclear to
me, but given that they did, the next table: 14 atoms of
carbon 1070.118 79.56 12 hydrogen 74.877 5.56 2 oxygen 200.000
14.88 -------- ------ 1344.995 100.00becomes perfectly clear.
For example, if one uses Berzelius' scale,wherein the weight
of oxygen is taken to be 100, the weight of carbonis about
76.44 and that is close to what one gets by dividing
1070.118by 14. So, the column on the left is simply their
calculation of theweights of 14 carbon atoms, of 12 hydrogen
atoms and of 2 oxygen atoms.Throughout the rest of the paper,
they use their own scale in which theweight of oxygen is taken
to be 10, and whenever they have 14 carbonsafter an analysis,
they give the weight as 107.0118, except for one placethat is
probably a typo. So, that explains the column on the left.As
for the column on the right, of one expresses these weights as
percentagesof the total weight 1344.995, one gets the column on
the right, more or less.So, now I have an overview of the kind
of computations they are presenting.What I'm still confused
about are their conventions for working with theirnumbers and
for dealing with questions of precision, etc. Note that
theytake all their numbers to three decimal places, even after
successivecomputations. Some of their results suggest that they
simply truncatedtheir numbers, no matter what came afterwards
(I think MF's lab instructorwould be pleased). But I don't
have it all nailed down.If, instead of 1.109 for the weight of
carbon dioxide, the value was really1.10891, and they used that
instead of the rounded value 1.109 in theiractual computations,
and if the weight of the oil of bitter almonds wasexactly
0.386, and if one used their value (according to Benfey's
footnote)of 7.644 for the atomic weight of carbon and 10 for
that of oxygen, then onewould get their value of 79.438 for
carbon in the first column of thecombustion analysis. I
haven't tried to find out all possible modificationsbut note
also that 1.1089 and 1.10892 won't work to do that. So, if
that iswhat they did, it would imply that they had a method of
weighing thathad a precision of 5 decimal places. Did they?So,
after all this math, I have a question about the chemical
laboratoryof Woehler and Liebig: how exactly did they weigh
things and what precisioncould they achieve with their
apparatus?I am cross posting this to sci.math, since someone
there might know somethingabout the history of numerical
methods in the time of Woehler and Liebigand have some idea of
what conventions they might have used in dealingwith numbers.
If someone can explain how to get EXACTLY the numbersWoehler
and Liebig claim to have derived from their initial
combustionmeasurements that would be very
helpful.Ignorantly,Allan
Adlerara@zurich.ai.mit.edu************************************
***************************************** ** Disclaimer: I am
a guest and *not* a member of the MIT Artificial **
Intelligence Lab. My actions and comments do not reflect ** in
any way on MIT. Moreover, I am nowhere near the Boston **
metropolitan area. **
**************************************************************
***************===
===
Subject:
: Re: How many ways to put 5 balls
into 500 ordered cups?Content-transferncoding: 8bitYou are
given 500 numbered cups and five identical balls. Any cup
canhold up to five balls. How many ways can you put the five
balls intothe 500 cups?> Lets make it more interesting: How
about *ten* balls (but each cup can still> only hold up to
five balls)?> -Michael.You want the number of integer
solutions toX_! + X_2 + ... + X_500 = 10 with 0 <= X_i <= 5
for i = 1 ... 500which is the coefficient of x^10 in (1 + x +
x^2 + ... +x^5)^500which is the coefficient of x^10 in (1 -
x^6)^500*(1 + x + x^2 + ....)^500which is C(509,10) -
C(503,4)which is 294,291,883,740,587,598,675-- Paul
SperryColumbia, SC (USA)===
===
Subject:
: To users of GMPThe FFT
code in GMP currently contains a flaw that leads to
inaccurateresults. The bug was discovered with the new
mpz_rrandomb routine earlier.It is recommended that users of
GMP recompile the distribution and use theoption
--disable-fft. More information can be found at
http://www.swox.com.===
===
Subject:
: Re: the anticlassicalist }{ i:
linguistic negationgalathaea> linguistic negationgalathaea> It
is not impossible to do that.Notice: my english, franch and
spanish knowledge is limited and possibly wrong.FYIin italian
(derived from Latin),double negation is grammatically correct
and commonly used,semantically sometimes is double meaning (1-
the double negationis considering as an enforcement of the
concept is being negated,2- the boolean not not x = x is just
only a possible interpretation.an example:non ho visto niente=
i didn't see anything (right whole meaning translation)but
literally (local mean, such as word by word / mean by mean)is
i didn't see nothingthe strange is that sometime the logical
meaning is biforked andtreated as a paradox, but it isn't
paradox, it's bad logicalgrammar definitionnon voglio
offendere nessuno= i want not offend someone (right whole
translation 1)literally:i would not offend nobody ( ~circa~
right translation 2)which means the opposite [if the last one
is not correct in english,then the meaning i tried to
translate was i want offend someone]All the two opposite
translations/interpretations are right.there is another
courious difference in logical keywords:with and without,
[economically abbreviated by w, w/o and w/?]in Italian
(with->con, without->senza) are the two unique
translations,the meaning of the two word is opposite (and
complementary), butsenza(whithout) doesn't contain con(with).
This mean that italianlanguage choosed to create a new word to
express an opposite concept.I noticed because.. Italian was my
first language, in that time...(my parent remember me) i used
the undefined con senza(wrong)in my first try to express
`without' concept)in spanish con / sin (similar to italian)in
franch avec / avec pas (or pas avec)Saluti,MARco mar.i.am--
x(t),y(t) =
th(3t-34.5)*e^[-(3t-34.5)^2]/2-4.3+e^(-1.8/t^2)/(.8*atg(t-3)+2
)(t-1.8)-.3th(5t-42.5),(1.4e^[-(3t-34.5)^2]+1-sgn[|t-8.5|-.5]*
1.5*|sin(pi*t)|^[2e^(-(t-11.5)^2)+.5+e^(-(.6t-3.3)^2)])/(.5+t)
+1 ; 0<t<14===
===
Subject:
: Re: Book on MATH game theory> theory in
the sense of economics at all, but a book that deals with >
actual games, and concrete mathematics. You could try Winning
Ways, by Berlekamp, Conway, and Guy.-- Gerry Myerson
(gerry@maths.mq.edi.ai) (i -> u for email)===
===
Subject:
: Re:
Teaching philosophyI can never resist diddling a pic or a
poem!!! &ante Melancholico ^^^^^^^^^^^^^^^^^^ To my darling
&rea & after all is said & done My life has been a lonely
one,& when I sometimes think of you I underst& your point of
view.& often whilst I contemplate, The time-worn s&s
accumulateBeneath my h& & fingers f& That ne'er gold b@ has
sweetly sp&===
===
Subject:
: Re: Request for comments on antiquated
algebraic topology online-book> Well I've pretty much decided
not to invest too much time with Lefschetz.> However,
initially, I only need to learn a small subset of algebraic>
topology (AT). Specifically, the same topics as found in the
first chapter> of Munkres' 'Elements of AT'. The application
is to distributed computing> where the evolution of finite
asynchronous distributed protocols are modeled> as high
dimensional simplicial complexes (if thats the correct
terminology).> combinatorial. The seminal explanatory paper on
the approach can be found> at:>
http://www.cs.brown.edu/people/mph/HerlihyR96/sv.pdfThat's an
interesting paper. I suppose the explanations are too brieffor
you? If you read that paper's explanations on topology and
combineit with looking at things online like Wikipedia, I bet
you'll bealright. See below also.> I have taken another look
at Hatcher, and you're right - it doesn't seem> nearly as
intimidating as it did before I bulked up on general topology
a> bit. Hopefully I won't need to shell out the cash for
Munkres' book.I think for you, it would be worth looking at
the first chapter ofMunkres. Just check it out from your local
university/college library. There is no need to pay up all that
money for something you only need apiece of.===
===
Subject:
: Re:
Integrals without explicit formulas> I am in a differential
calculus class. Just about all the problems> involve
integrals, but I have been informed that some of the
integrals> in these problems have no explicit formulas. Can
you give me any rules> of thumb so I know what kind of
integrals to not bother trying to> integrate?Instead of no
explicit formula I would say no closed form expression in
elementary functions, but I think we're talking about the same
thing. I don't think there are any rules of thumb for what you
want. Functions that have no elementary antiderivative can
look a lot like functions that do. e^(x^2) has none, xe^(x^2)
does. log x does, 1 / (log x) doesn't, 1 / (x log x) does, x /
(log x) doesn't, (log x) / x does. There is a way to work out
which ones do & which ones don't, but it's a bit beyond the
level of an intro calculus course.-- Gerry Myerson
(gerry@maths.mq.edi.ai) (i -> u for email)===
===
Subject:
: Re: HELL
:-) (Was: Re: the anticlassicalist }{ ii: the spectre
continues)*MATHEMATICS IS FULL OF TRICKS*> French even allows
you the conjugation :> ma th.8ematique> ta th.8ematique> sa
th.8ematique> notre th.8fme .88 tics> votre th.8fme attique>
leurre, t'es mat, hic !*REDUCING TOPICS TO THE SCIENCE OF
NUMBER IS NUMEROLOGY*> I'd call that accounting.Speaking of
accounting, however,... it has this tendency to rely
ondoublentry bookkeeping. That is more than can be said for
people whodraw pictures on a blackboard and apply a
calculus.Apparently, Robinson supplied a concrete sense for
infinitesimal. Ilooked at one of his papers, and it seems to
involve a quotient ofrings. Should the many-valued
interpretation for quantum mechanicssurprise anyone if the
history of the mathematical development isignored? Of course,
the information media is running around tellingpeople that
they are one of infinitely many dopplegangers all of whomare
multiply-situated in as many parallel universes as
eventpossibilities since the beginning of time.Talk about
failing to grasp the import of Cantor's
paradise.:-)mitch===
===
Subject:
: Generalization of Cauchy's
Functional EquationThe continuous solution of Cauchy's
functional equation,f(x+y) = f(x)+f(y)is f(x) = cx. Then what
is the solution of the generalized Cauchy'sfunctional equation
such as,f(x+y) = f(x)+f(y)+a{f(x)f(y)}^2 ?I tried to solve the
generalized Cauchy's functional equation in thefollowing
site:http://139.134.5.123/tiddler2/cauchy/
cauchyequation.htmand also tried to simplify the addition
theorem for Weierstrass ellipticfunction.===
===
Subject:
: Re:
RFI:Wanless' Fourth Conjecture/Dirichlet's Geometric Theorem>
The series a^n+b [hcf(a,b)=1] includes an infinite number of
primes [over> all n]The sequence 2^n + 78557 includes no
primes.-- Gerry Myerson (gerry@maths.mq.edi.ai) (i -> u for
email)===
===
Subject:
: Re: JSH: Making it personal> I'm concerned
about all the histrionics that has darkened even my own> posts
as it's all too personal at this point.> Right now I *despise*
posters like Dik Winter, Nora Baron, and David> Ullrich, which
necessarily involves some fear: they're potent at> fighting for
their positions.> They also just so happen to ultimately be
wrong.Whereas JSH is merely continuously and inevitably
wrong.> Dik Winter keeps up webpages on some of my OLD failed
attempts at> proving Fermat's Last Theorem.> That's just bad
behavior.Not nearly as bad as your posting and reposting all
those errors in the first place.> Trotting out someone else's
mistakes made years ago, and pushing them> on your own
personal webpages is not something that's defensible.> It's
clearly making a statement.It is a warning to the innocent
that there are liars around.> Dik Winter invites my contempt,
and unfortunately, I have to fear the> effectiveness of such
tactics.I think it is more that Dik is expressing than
inviting comtempt.> It's sad really.You are indeed.> I'm going
to see about dropping that fear as I think it gets in the> way
of research, but right now I do despise these people because>
they're so contemptible, and so successful in screwing up the>
dialogue.It is the monologue of falsehoods by JSH that has been
sucessfully screwed up. And that is admirable, at least to
everyone exceot JSH.===
===
Subject:
: Re: Graph Theory Textbook> I'm
an undergraduate, and this summer I will be participating in
an> REU for discrete math and combinatorics. I am looking for
a good> graph theory textbook to learn the basics from. I have
seen the books> that Dover (which of course are very cheap) has
to offer on the> subject, but unfortunately they seem too
elementary. Is the Springer> GTM Graph Theory by Reinhard
Diestel any good? Any suggestions are> highly welcomed. There
is a Dover book by Claude Berge on Graph Theory which is
definitely not elementary, even for an experienced student in
graph theory.Berge's Dover
book:http://www.amazon.com/exec/obidos/tg/detail/-/0486419754/
qid=1077169399/sr=1-1/ref=sr_1_1/104-7198216-1259942?v=glance&
s=books An old book good for introductory - yet serious and
real - graph theory is the textbook by Bondy and Murty, but
this might be a little old and not too available, but I've
heard they have created a new edition. Of course, there are
many fields of graph theory... if you are approaching it from
the computer science point, you might like Algorithmic Graph
Theory... by Martin Golumbic.J===
===
Subject:
: combitoricsI'm not
sure if this is a good group to post this.I have a
combinatorics problem. well, a puzzle really. I'm not surewhat
kind of problem this is. so I would love to get some
feedbacksfrom the group. ( as to the nature of the problem as
well as how tosolve it)Let's say you have a little sister
named Torrey. And she has quite aquite a number of friends in
kindergarten. let's say her friendshipgoes as follows.Torrey
loves Matt on some days, but Torrey hate Matt on other
days.Torrey also loves Paul on some days, but Torrey hates
Paul on otherdays.Torrey also loves Sandy on some days, but
she never hates Sandy.Sandy hates Torrey on some days, but
loves her on other daysMatt loves Sandy on some some days, but
hates her on other other daysPaul Loves Matt on some days, but
Paul hates other days.paul hates Sandy on some days. to make
it a bit simpler...Torrey loves and hates MattTorrey loves and
hates PaulTorrey loves Sandy Sandy loves and hates TorreyMatt
loves and hates MattPaul loves and hates MattPaul hates
Sandyhow can I figure out how many possible combinations there
are thatdesribes the love and hate relationship between Torrey,
Matt, Paul,and Sandy?I have 32. but I'm not sure if i'm right.
and i'm doing it brute forceway.any helpful thoughts and
comments are appreciated. thank you all. sean===
===
Subject:
: Re:
the anticlassicalist }{ vi: into the quantum-=-=-=-=-=-= into
the quantum =-=-=-=-=-=-=-Before you lies the Void. Some use
the word vacuum here, but the vacuum ismore rigorously used as
a term of a particular subspace structure I willdescribe. Some
might call it space, but that too is structure, and I wantto
address something prior to structure.I want to address
ontology.These are the words we build our models of reality
with.If I haven't made it clear enough yet, this is my most
fundamental concern.When I think about the world, I want to be
precise. I want to have a skillwith the concepts. I feel many
others here feel exactly the same. We enjoylearning or have
some drive to learn because all these groups
discussknowledgable things.Maybe I am just naive and young,
but I see very smart people, some who liketo show off, some
who teach, an annoying gang of aggresive types, some veryhere
are interested in such things. It is usenet among the crowd
thatinterprets ontologies in a rigorous science of
symbology.And you all have models, built on ontologies
possessing vary many logics.We even use them in everday speech
or natural language. We order ourspeech to be useful.The
ontology is where we define the objects and their
transformations. Itis where we define a language about which
we talk.We build symbolic names for the patterns we recognise.
These naturallyclassify themselves into objects and
transformations.Let me introduce the ontology of the
quantum.In models, we start with a structure over an
ontology.A structure is: - a collection of names for objects
:: both directly associated to existents and allowed :: :: to
vary, freely and bound :: - a language of truth {quantifiers,
(co)products, negation, implication,etc.} - a collection of
operations taking collections of objects back into theobject
domain - a collection of relations mapping collections of
objects into the truthdomainMathematicians use structures all
the time. Magmas, semigroups, groups,rings, fields,
topologies, orders, etc. But it is very likely moreuniversal
than that. As I've tried to show with the building of
thecognitive maps, the ontology of natural language and our
ability torecognise objects and transformations in the world
abstractly through alanguage, can be structures as well.Our
quantum ontology is built inside a structure known as a
Hilbert space.I will not describe the full definition, but if
you are familiar with linearspaces, like Euclidean space,
studying Hilbert spaces is very much the studyof what
generalises when we look to include infinite dimensions in a
linearway, ie. what is common to the finite and infinite
spaces with a certainbinary form that makes the space
metrically complete.Hilbert spaces have a beautiful geometry,
much inherited from their parentBanach spaces but with added
structure. And this is the ontology quantummechanics first
settled in on (there have been numerous extensions, like
thebeautiful geometric extension of B. J. Hiley's Algebraic
quantum mechanics,algebraic spinors, and Hilbert space which
looks to Heisenberg algebras andhas some ties with realist
interpretation theory).State is a vector or ray in a Hilbert
space. We build this to anepistemology through a language of
observables. The observation mechanismincludes applying
projection operators to the state to get a mapping intothe
interval [0, 1], from which our epistemology gets a
probabilisticinterpretation.Each projection operator
corresponds in a natural way to a closed linearsubspace of the
Hilbert space through the operator spectrum structure of
thespace.So when we want to be able to discuss making logical
propositions of eventswith conjunctive and disjunctive
connectives, negation, and general build asentence structure
to be able to make predictions of a logical kind forempirical
review.The way this operates in quantum mechanics is by
looking at the naturallattice structure of the closed, linear
subspaces under the identification:[A is a subspace of B] =>
[Lattice(A) -> Lattice(B)]Here meet and join, product and
coproduct, etc. can both defined naturally,in the same way I
mentioned previously, and identified with closed spans
ofunions and intersection of these closed linar subspaces.With
these notions, it is possible to form experiments looking for
theprobability of event A and event B and event A or event B.
You can findthe lattice element corresponding to the logical
structure of a sentence andproject it into a probability which
you can compare to experiments and theirabstractions.There is
even a natural negation that corresponds to the
properepistemological predictions of the theory through what
is calledorthocomplementation of subspaces, where you look at
the spans where innerproducts become zero.derived.
Orthocomplemetation ~() obeys:~(~(a)) <-> aif a -> b, then
~(b) -> ~(a)(a / ~(a)) <-> Lattice(0)(a / ~(a)) <->
Lattice(I)where Lattice(0, 1) are the _|_, T respectively of
the lattice. Besides thecommon relationships for / and /
general to lattices, there isadditionally the relation(a / (b
/ ~(a))) <-> bwhich describes how negation on one subspace can
be used to separate asecond subspace into a section whose
conjunction with the former recoversthe latter. The
probability of one event does not change if we look atrelated
to another event and this latter event's opposite in a
correlationexperiment.This defines a logic that is called
orthomodular (for finite dimensionalHilbert spaces it is fully
modular, ie. (a / (b / c)) <-> ((a / b) /c). It is the logic we
build epistemological statements with in order tocompare with
experimental observation. If you were to try to use
booleanconnectives here, you would give predictions that
differ from experiments.Most physicists don't pay too much
attention to this, because they'velearned the formalism to
convert a problem into the language of algebra,often without
being pointed to the logical structure that defines
thealgebra, but that is the structure of reasoning in a
quantum ontology.There are natural connectives as well for
implication and quantifiers.It is interesting to note that a
difference between Hilbert space and thisstructure uniting
ontology with epistemology for the quantum model andclassical
mechanics in Euclidean spaces is captured by the geometric
notionscontained in the boolean and quantum logics. For one,
operators need notcommute.Now, there is a very interesting
theorem out there which proves:There exist no partial Boolean
algebra homomorphism from the lattice algebraof a Hilbert
space into a Boolean algebra whenever the Hilbert space is
morethan 2 dimensional (many Hilbert spaces for quantum
systems are infinitedimensional).So, models with non Boolean
structure are fairly necessary to the study ofquantum
mechanics. As my other posts have tried to underline, models
areeverywhere that use nonclassical logics directly. There are
many modelswhere the epistemology is connected to the ontology
through logics that cannot always derive true or false.And
what I mean by relating a sense of absoluteness (like a form
ofPlatonism) to this belief that all logic should be viewed as
Boolean is thatit states often in some form that all
propositions have a true or false evenif we cannot derive the
answer. But that is actually a provablyinconsistent statement
for quantum models. Even John von Neumann's originalno go
theorem for hidden variables in quantum mechanics actually
pointedto the lack of a probabilistic embedding.Sure, the game
you play with the symbols to understand the various
logicsdescribed by many models can be taken to be many things.
That's studyingsomething completely different, and I've shown
that some cognitive researchwould point to a different logic
being more faithful.I haven't mentioned |= yet.I am not
satisfied...--
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-galathaea:
prankster, fablist, magician, liar===
===
Subject:
: Re: No Set
Contains Every Computable Natural
<O1xXb.12743$B53.5007@newssvr29.news.prodigy.com>
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<M5adnQCN-sgwUK7dRVn-ig@comcast.com> Discussion, linux)> I
don't know what it means for N to be computable. I think it's>
pretty clear what I mean when I say N is a recursive set.
Evidently,> it's not clear to you, but maybe the above helps.>
Given any set, S, of representations of natural numbers,> I can
write a TM that will find a representation of a natural number>
that is not in S. (No, this TM does not halt after a finite
number> of steps.)This is, of course, simply false, false,
false.If you give the set of all representations of natural
numbers, thenyour TM will not find any representation of a
natural number not inyour set.You're speaking nonsense
again.Of course, we must be careful here and fix a convention
for when atape represents a set of natural numbers. Let's say
that a taperepresents a set of natural numbers if it consists
of blocks of onesseparated by a single space.We must also be
careful to require that your TM has the property that,for each
square x of the tape, there exists a steps t such that for
allsteps s after t, the symbol on the square x at time s is
the same asthe symbol on x at t. After all, a TM which
repeatedly changes the0th square to 1 and blank and 1 and
blank cannot be said to produceany tape at all.Now, on an
input tape containing 1, ,1,1, ,1,1,1, ,1,1,1,1, ,...,there is
simply no possibility that your TM, even after an
infinitenumber of steps, creates a tape with a representation
of some naturalnumber not on my input tape. Every natural
number is represented onmy tape. None is missing. Your tape
either returns a set of numbersalready on my tape or it
returns no set of numbers at all.> Obviously, a human has to
decide if a symbol represents> a natural number. No TM can do
this.No TM can devise a convention, if that's what you mean.
At least, noTM can devise a convention in the sense that
matters here.> But, a TM can extend the convention in such a
way that no set> contains every symbol that represents a
natural number.No, no, no. A TM cannot extend the
representation of natural numbersfor two reasons: (1) It's a
ing Turing machine, isn't it? I don'twant to argue about
requirements for intentionality or whether TMs arecapable of
intelligence, but this TM is not part of the
negotiationsregarding our conventions. (2) Every natural
number has arepresentation in our convention, so the
representation cannot benon-redundantly extended.You really
aren't thinking very clearly.-- Come on people!!! The US just
blew up a lot of people in Iraq, don'tyou realize that a
person with my exposure might just end up dead, bymysterious
circumstances? --James Harris, on the dangers of proving
Fermat's last theorem===
===
Subject:
: Re: Advice on future with
Math> I'm an upperclass math major that was/is planning on
attempting a> masters or phd in math. Here lately, I've
wondered if this is a great> idea. > My first question is
this: How much should you study (reading the> material,
working problems) for, say, an abstract algebra or advanced>
calculus junior/senior level class? I talked to a graduate
student> that said he studied for a couple of his first year
grad classes at> least 4 hours a day on weekdays and about 8
hours on Saturday and> Sunday. Needless to say, I do nothhing
like that. Should I be doing> that or is that just an insane
amount of time to be spending on it? > It seems that if you're
good at something, it shouldn't require so> much work. I might
be an anomaly, but I don't spend much more than 2 hours a day
actually sitting down with pencil and paper working on math.
However, what I am doing is almost every minute I am awake and
every single minute I am asleep, I am thinkingabout math, or
one of the problems I am working on. Occasionally I'll grab
some scratch paper and jot something down. Most of the time,
not.Okay, maybe saying every minute is a little exaggeration,
but not much. Seriously, though, I spend the best part of my
day, everyday thinking about math. If I am watching TV or
reading the newspaper, or talking with someone, in the back of
my head is something like, What is the solution to the PDE
a*u_x + b*u_y + c*u_z = 0?Also, it has been my experience that
if you have to spend insane amounts of time studying math, then
maybe continuing on is not for you. I am not trying to
discourage you, and I would hope you at least try for a year,
but if you are stuggling and spending 8 hours a day in the
books, you might want to look for something else.Also, there
are many different branches of math. Personally, I am of the
discrete/combinatorics and numerical methods persuasions.
Analysis, topology and PDE, I only took those classes becaues
they were required. And, I struggled through them.> At times I
have difficulty understanding things that, after finding> out
where I went wrong, seem pretty simple. I assume others don't>
have this problem because normally they can respond right away
with> what some math concept meant or, less often, help me
where I am having> a problem understanding part of a proof. Is
it normal to not follow a> proof or just a sign that I'm just
weak mathematically (we're again> talking junior/senior level
advanced calc/abstract algebra here)?If after finding out
where you went wrong things are still difficult, you are,
well, screwed. The right answers always look easy. Then you
think, Who the heck thought of doing it that way? However,
that is not to say you shouldn't be able to follow a proof. I
read many papers and I can't always follow the proofs givne in
them. Often, they skip steps or say somethign is obvious when
it isn't (unless you know som obscure theorem). But, if the
proof is in your text, you should, for the most part, be able
to follow it.> I have a bad tendency to be hot/cold when it
comes to math. It seems> like I'll have the highest/near
highest score on an exam and then> really screw up on the next
exam, getting something in the bottom half> of the grades. I
wish that I could attribute it to something like> when I first
started school: I would do no work until after I screwed> up on
the first exam, and then work harder after that and do quite>
well. Now it seems that I'll do well on the first exam and
then screw> up on the next one. I really don't think that I
will have studied > much less for the second one but I guess
it could be a possibility. > When I do bad on the other exam,
I begin to wonder if I just got> lucky on the problems that
were given on the good exam and would have> done lousy if a
few different problems were picked.You have to remember that
all math is comprehensive. You couldn't do modern algebra if
you didn't know how to add and multiply. The biggest thing
that I have found that helped me was relating one subject to
another. This is especially easy when the subjects are closely
related, like topology and analysis. But try it for subjects
that seem less related. The triangle inequality in linear
algebra is the (exact) same one you learn about in functional
analysis. Once you start doing this, it becomes easier to
learn, because you aren't learning all new stuff, you are just
review things, popssibly with different names. The solution to
the heat equation is just an inverse problem, but using
forawrd modelling (which makes it easier).> Other times, I
wonder if I am just not good at proofs, thus making me> about
the crappiest grad school candidite out there. I sit and
stare> at the homework problems and often can't get anywhere.
After I see an> answer, it doesn't look too bad but it just
seems that they're> impossible at times. I had a bad proofs
class that was designed for> just about anyone wanting to take
it, so it was really easy. I'm not> sure if I can use that
excuse at this point though. My pre-calculus> education was
really bad since I was from a really poor school (I > often
have no clue what certain things that were assumed as>
'knowledge' from high school) but again this doesn't seem to
matter> too much in the proof-based classes.Proof writing
takes skill, alot of skill. It also take practice, even more
practice than skill. But, the most important thing of all is
that writing good proofs require you to have seen well written
proofs. If you are taking modern (abstract) algebra, go to the
library and check out a few other books on it. Start looking
the QA section. Often times, different books will have very
different proofs for the same theorems. Often times, in one
book they prove theorem A and use that to prove theorem B,
then another book proves theorem B and uses it to prove
theorem A. Try looking in a few different analysis books for
the proofs of the following theorems:Axiom of Choice (okay,
this is a given)Nested Interval
PropertyBolzano-WierestrassCauchy CriterionMontone Convergence
Theorem.All of these theorems are equivalent (more or less) so
the order they are proven, and which is used to prove the
other, is very much up to the author. In fact, on an analysis
test, one of our questions was to prove one of the 4 theorems,
assuming a different one is true, not the same way the text did
it. Our text did AoC => NIP => BW => CC and AoC => MCT. I think
I showed CC => MCT on the test> Well, that was kind of
long-winded. I hope that I got the point> across. Any advice
would be helpful. Okay, those are my thoughts. Take them and
do what you will with them. But, I tyhink if you reall ylike
math, and are not a burnt out senior who is just looking
forward to graduation, you should at least try a semester
working towards a master degree. If you find out it's not for
you, well, you learned and you didn't really waste any of your
life. But, if you don't try, you will always be kicking
yourself. - Tim-- Timothy M. BrauchGraduate StudentDepartment
of MathematicsWake Forest Universityemail is:news (dot) post
(at) tbrauch (dot) com===
===
Subject:
: Re: Metamath Axiom of
Choice> ...................>As I stated above, the Metamath
formulation of the Axiom of Choice is >effectively the
statement that given any set x, there exists a set y >such
that for all nonempty elements w of x, w n (Us) has exactly
one >element, where s = {t in y : w in t}. >As stated, it is
false. Add the elements of x are disjoint>and it is one of the
standard forms.>To see that it is false as stated, let x have
three elements, >{1, 2}, {1, 3}, {2, 3}. Any set intersecting
all of these in>at least one element has to contain both
elements of one.>But in the statement above, s is explicitly
dependent on w (the >is not the existence of a single set with
one element in common>with every nonempty element.In view of
possible misinterpretations of the Metamath statement of the
Axiom of Choice, the statement is effectively the statement
thatfor all sets x, there exists a set y such that for all
elements w of x, if w is nonempty, then w n (U{t in y : w in
t}) has exactly one element. At the moment, they (the Time
Lords) are far from being all-powerful. That's why it's been
left up to me and me and me. quote by: Patrick Troughton in
The Three Doctors-------===
===
Subject:
: Re: combitorics> Torrey
loves and hates Matt> Torrey loves and hates Paul> Torrey
loves Sandy> Sandy loves and hates Torrey> Matt loves and
hates Matt> Paul loves and hates Matt> Paul hates SandyWell,
there are five loves and hates relationships in the
abovedescription. Assuming that they are independent, then the
total number ofpossibilities are 2^5=32.===
===
Subject:
: Re: looking
for an eltry soln to an old problem> A long time ago, J.J.
Sylvester posed the problem:> if I have arbitrarily many 5
cent or 17 cent stamps,> what is the largest denomination I
cannot make?> In general, if we have p and q cent stamps, it
turns> out the answer is pq-p-q (granted p and q are coprime).>
I have derived a solution to the problem, but I'd like> to
teach this to my undergraduates, some of whom have a> limited
background. So my question is: is there a very nice> &
friendly proof of this fact? (For example, which avoids> any
nonobvious facts from number theory.)> You need a couple of
not-quite-obvious facts.> Fact 1: if x = a p + b q, then x =
(a + n q) p + (b - n p) q for> any integer n, and these are
all the ways of writing x as a> multiple of p plus a multiple
of q.> Fact 2: if p and q are coprime, every integer x can be
written as> x = a p + b q for some (not necessarily positive)
integers a,b.> This can be done using the Euclidean algorithm;
or the fact that> t -> t p mod q is one-to-one on the integers
in {1,...,q-1} coprime> to q, plus the pigeonhole principle.>
Suppose for some x it can't be done with a, b >= 0. If> x = a
p + b q is one representation with integers a,b, then> for
each n we have either a + n q < 0 or b - n p < 0.> Take a' to
be the largest a + n q < 0, so -q <= a' < 0 and> x = a' p + b'
q with b' - p < 0.> But then x <= -p + (p-1) q = pq - p - q.>
On the other hand, pq - p - q = ap + bq with a=q-1 and b=-1;>
a+nq < 0 if n < 0 and b-np < 0 if n >=0.Clearer: we may
normalize N = P X + Q Y so 0 <= X < Q by adding a certain
integral multiple of (-Q,P) to (X,Y)CLAIM: N = P X + Q Y for
some integers X and Y >= 0iff its normalization has Y >= 0.
PROOF: X and Y >= 0implies normalization requires addition of
(-Q,P) zeroor more times, and this preserves the condition Y
>= 0.Conversely if the normalization has Y < 0 then N has
norepresentation with X and Y >= 0, because to shift Y >
0requires adding (-Q,P) at least once, which shifts X <
0.Finally, since X P + Y Q is increasing in both X and Y,it is
clear that the largest non-representable number Nhas
normalization (X,Y) = (Q-1,-1), so N = PQ - P - Q.Notice that
the proof has a vivid geometric picture: representations of N
correspond to lattice points (X,Y) on the line N = P X + Q Y
with negative slope = -P/Q.Normalization is achieved by
shifting forward/backwardalong the line by integral multiples
of vector (-Q,P)until you land in the normal strip where 0 <=
X < Q-1.> In case you're interested, the class I'm teaching>
is linear algebra, but as you can see I like to > give puzzles
to the class which are not necessarily> related to the material
(in an obvious way).Here the underlying linear structure is a
Z-module,a generalization of a vector space; i.e. here the
scalars are the integers so have only the structure of a ring,
not a field. Unless you've already taughtsome module theory, it
might be tricky to preciselyexplain the relationship to vector
space theory.Finally it should be mentioned that there has
been muchwritten on this classical problem. To locate such
workyou should ensure that you search on the many aliases,e.g.
postage stamp problem, Sylvester/Frobenius problem,Diophantine
problem of Frobenius, Frobenius conductor,money changing, coin
changing, change making problems,h-basis and asymptotic bases
in additive number theory,integer programming algorithms and
Gomory cuts,knapsack problems and greedy algorithms, etc.-Bill
Dubuque===
===
Subject:
: Re: Teaching philosophy> To &rea (names have
been changed to protect the guilty)Ampersandrea? That's sure an
unusual name. :-)Consider saying something about the kind of
questions youlike to ask your students. What do you do to keep
yourstudents from falling asleep? What kind of motivation
haveyou had the best success at? Best of luck; I would
havefound this difficult when I was applying for such
jobs.Keith Ramsay===
===
Subject:
: Re: Teaching philosophy> To &rea
(names have been changed to protect the guilty)Ampersandrea?
That's sure an unusual name. :-)Consider saying something
about the kind of questions youlike to ask your students. What
do you do to keep yourstudents from falling asleep? What kind
of motivation haveyou had the best success at? Best of luck; I
would havefound this difficult when I was applying for such
jobs.Keith Ramsay===
===
Subject:
: Re: Teaching philosophy> To &rea
(names have been changed to protect the guilty)Ampersandrea?
That's sure an unusual name. :-)Consider saying something
about the kind of questions youlike to ask your students. What
do you do to keep yourstudents from falling asleep? What kind
of motivation haveyou had the best success at? Best of luck; I
would havefound this difficult when I was applying for such
jobs.Keith Ramsay===
===
Subject:
: Re: No Set Contains Every
Computable Natural (was Church-Turing compared to Zuse-Fredkin
thesis)messagemessage> This is the definition of a
recursively enumerable language:> This language is
recursively enumerable. Given a string, x, thereexists> a> TM
that> will halt after a finite number of steps if x is a
member of L.> This language is NOT decidable. Consider what
happens if x is an> infinite string of 1's. The TM will not
halt after a finite number> of> steps.> You can't input an
infinite string to the tape. If you could, itwouldn't be a
Turing machine.> No, but you can input finite instructions
which generate infinite> output.> Russell is right about
Turing saying this, and I found it repeated:But that would be
a different language. The language would then be {<M>|> Mis a
TM that outputs a string a finite string of 1s}. This language
isclearly undecidable. But that does not prove his point, as
any languagewhere (language U lang_complement) is composed of
arbitrary TMs andwhosemembership is based on determining
non-trivial properties of those TMs,isundecidable (a la Rice's
theorem). However L is certainly decidable:<quote>Let me define
a language, L, that consists of all unary
representationsofnatural numbers.1 = one, 11 = two, 111 =
three, etc.</quote>l8r, Mike N. Christoff> You can't input an
infinite string to the tape. If you could, itwouldn't be a
Turing machine.> Harris:> No, but you can input finite
instructions which generate infinite> output.> Russell is
right about Turing saying this, and I found it repeated:>
Chaitin et al: This is a machine that performs an
unendingsymbols.> [SH: If you read the thread, Russell
previously claimed this.}> Harris added in last post:I posted
not to argue on Russell's side, but because I thought this>
Turing ingenuity deserved some recognition.I didn't get that
far. You mention this at the very end, you should
havementioned it at the beginning. I assumed you were
equivocating Russell'sinfinite input on a TM input tape claim
with the finite instructions onthe tape idea.As to your
original point, I agree that not-halting or looping caneither
mean repeating a sequence of states forever, or taking an
unendingnumber of steps while never repeating a
state.<muse>I'm remembering Wegner's interaction machines.
This may be a complete liftof Wegner, but here's an idea.
Define a so-called 'stream' TM that can reada possibly
infinite input stream and produce a possibly infinite
outputstream. The TMs being discussed would then be
finite-input/infinite-outputstream TMs. So instead of
'non-halting', you would say 'streaming'.However, a stream TM
would have both an internal tape, and a write-onlyoutput tape.
Any TM that can (*)-output its stream to any finitespecified
length, in finite time, can be modified to use a seperate
tapefor output in an obvious way.To be a stream TM a machine
must have one or both of : infinite output orinfinite input.
Stream TMs can be either infinite or finite state. The
TMdiscussed would have to be infinite state. Any finite state
infinite-outstream TM has an output that eventually repeats
forever, therefore itsoutput set must be countable; each
element representable by a rational. Asa generalization of
Turing's rule (*), I propose : (**) for every finiteprefix O
of the output, there exists a finite prefix I of the input,
suchthat O appears on the output within a finite number of
steps after I isread.In the case of a stream language L, (and
I definitely recall reading aboutthis in a Wegner paper), we
can define L: input -> output as L: R -> R, atleast for some
inf-in/out infinite state TMs. But where Wegner gets itwrong
is to imply that since we seem to be mapping reals to
reals,interaction machines are superTuring. What he fails to
realize is thatalthough the cardinality of the languages
recognized by these machines hasgone from countable to
uncountable, we have not upgraded our definitionstating a
computation must take a finite number of steps. (Note, it
makessense to replace 'steps' with 'time', since steps implies
a particular modelof computation). Maybe call this notion limit
computable, or streamcomputable. For example, although it may
be able to stream-compute pi, astream TM is not superTuring
unless it could output pi in finite time.Languages would
probably end up being defined in terms of the output
setinstead of the input set, since one can no longer accept or
reject an inputfor any inf-in stream TM. This would fit
perfectly in terms of TMs thatstream compute numbers like pi,
since the important stuff is in the outputnot the
input.</muse>Hopefully that did not seem too 'crank-like'.
Also, I will reread Wegnerand see just how badly I've
plagiarized :) But one can see how this ideacan go in many
many directions. For instance, stream TMs that can read
andapply new code replace old code. ie: extending the state
space withoutneccessarily increasing memory usage for long
lived processes. Also I'llprovide a link, if anyone's
interested, to the Wegner papers(s). (btw: asearch for, wegner
interaction machine, should find his stuff)l8r, Mike N.
Christoff===
===
Subject:
: Re: Teaching philosophy> I need to write
a description of my teaching philosophy. In order to> do so
accurately, I think I need to write at greater length and in>
greater detail on this topic than any hiring committee will
want to read.> Moreover, I need to present my views accurately
but, somehow, in such> a manner as not to vitiate the
consideration of my application.> That being the case, can
someone please tell me what my teaching> philosophy is?> What
kind of philosophy? Metaphysics? Epistemology? Ethics?
Aesthetics?> Kantian? Pragmatic? Platonic?
Aristotelean?Cynical? (the school of Diogenes).-- ===
===
Subject:
:
Re: intercept length when a random line intersect with
ellipsoidDear Mathman,The general question is that if a
infinite plane with any orientationintersect with a polyconvex
shell with the fixed thickness(A), thethickness of intersecting
part is larger than A. So I want to know howmuch is the average
thickness of the intersecting part from theviewpoint of
probability? Since I'm not sure the problem can be sovledfor
any polyconvex shell. So I select the simple shape, like
ellipsoidshell or orthogonal parallelepiped shell with a fixed
thickness.Now I want to know, do we talk the same
question?Thank you for your kind reply again!CHEN Huisu> In
order to answer your question, you need to be more specific
aboutrandom for the line. A line in space can be specified as
all point> P satisfying P=X+tV, where X is a random point in
space (needing a> probability distribution) and V is a random
unit vector (isotropic?> distribution), while t ranges over
the entire real line.> Once these distributions (for X and V)
and the parameters of the shell> are defined, you will have a
tractable porblem.> Posted Via Usenet.com Premium Usenet
Newsgroup Services>
----------------------------------------------------------> **
SPEED ** RETENTION ** COMPLETION ** ANONYMITY **>
---------------------------------------------------------- >
http://www.usenet.com===
===
Subject:
: Re: the anticlassicalist }{
i: linguistic negation> In message
<1035esdt30h4d38@corp.supernews.com>, galathaea>Of course, you
also snipped all the references to math and physics,> Did I
blink? I haven't seen any physics yet.You should have realised
that anything full of -ism and -ist which is spewedto sci.lang
and sci.logic is not going to contain any
physics.Franz===
===
Subject:
: Re: RFI:Wanless' Fourth
Conjecture/Dirichlet's Geometric Theoremafter [though I'll
admit that I was hoping there wouldn't be any, for
theConjecture's sake!]I'll investigate and double-check in due
course [not that I don't believeyou! :-)]JThe series a^n+b
[hcf(a,b)=1] includes an infinite number of primes[overall n]>
The sequence 2^n + 78557 includes no primes.> --> Gerry Myerson
(gerry@maths.mq.edi.ai) (i -> u for email)===
===
Subject:
: Re:
RFI:Wanless' Fourth Conjecture/Dirichlet's Geometric
TheoremA)Even I can't prove something that isn't true!!!
:-)B)I'm assuming you're claiming that for the values of a and
b you quote,there are _no_ primes in the sequence.So thanks for
this David, it's exactly the sort of information I wasafter
[though I'll admit that I was hoping there wouldn't be any,
for theConjecture's sake!]I'll investigate and double-check in
due course [not that I don't believeyou! :-)]J_If_ it's true -
can one prove it from Dirichlet's [Arithmetic
iestandard]Theorem?> A) Regardless of its truth or falsity, I
am sure that you can prove it> from Dirichlet's Theorem.> B)
Consider a=234578273658273645827341
b=2348798769847962341J>The series a^n+b [hcf(a,b)=1] includes
an infinite number of primes[over>all n]===
===
Subject:
: Re:
RFI:Wanless' Fourth Conjecture/Dirichlet's Geometric
TheoremA)Though, of course, (you or) I might be able to
disprove it... :)))> A)Even I can't prove something that isn't
true!!! :-)> B)I'm assuming you're claiming that for the values
of a and b you quote,> there are _no_ primes in the sequence.>
So thanks for this David, it's exactly the sort of information
I was> after [though I'll admit that I was hoping there
wouldn't be any, for the> Conjecture's sake!]> I'll
investigate and double-check in due course [not that I don't
believe> you! :-)]> J> _If_ it's true - can one prove it from
Dirichlet's [Arithmetic ie> standard]> Theorem?A) Regardless
of its truth or falsity, I am sure that you can prove itfrom
Dirichlet's Theorem.B) Consider a=234578273658273645827341
b=2348798769847962341> J>The series a^n+b [hcf(a,b)=1]
includes an infinite number of primes> [over>all
n]>===
===
Subject:
: Re: No Set Contains Every Computable
NaturalGiven any set, S, of representations of natural
numbers,I can write a TM that will find a representation of a
natural numberthat is not in S. (No, this TM does not halt
after a finite numberof steps.)> This is, of course, simply
false, false, false.> If you give the set of all
representations of natural numbers, then> your TM will not
find any representation of a natural number not in> your set.>
You're speaking nonsense again.> Of course, we must be careful
here and fix a convention for when a> tape represents a set of
natural numbers. Let's say that a tape> represents a set of
natural numbers if it consists of blocks of ones> separated by
a single space.> We must also be careful to require that your
TM has the property that,> for each square x of the tape,
there exists a steps t such that for all> steps s after t, the
symbol on the square x at time s is the same as> the symbol on
x at t. After all, a TM which repeatedly changes the> 0th
square to 1 and blank and 1 and blank cannot be said to
produce> any tape at all.I have given this proof several
times.My TM has the property you describe.> Now, on an input
tape containing 1, ,1,1, ,1,1,1, ,1,1,1,1, ,...,How about
b1b11b111b1111...> there is simply no possibility that your
TM, even after an infinite> number of steps, creates a tape
with a representation of some natural> number not on my input
tape. Every natural number is represented on> my tape. None is
missing. Your tape either returns a set of numbers> already on
my tape or it returns no set of numbers at all.This TM will
find a representation not on your tape:It is a three state
machine and I can provide a statetransition table if you
like.1) Find a blank2) Find a second blank3) Backup and write
a 1 on the previous blankrepeat steps (1) through (3)This TM
will produce a tape that contains exactly one blank.The
contiguous string of 1's preceding this blank will bea
representation not on your tape.(Others have noted this is a
very crude adder.)> Obviously, a human has to decide if a
symbol represents> a natural number. No TM can do this.> No TM
can devise a convention, if that's what you mean. At least, no>
TM can devise a convention in the sense that matters here.But,
a TM can extend the convention in such a way that no
setcontains every symbol that represents a natural number.>
No, no, no. A TM cannot extend the representation of natural
numbers> for two reasons: (1) It's a ing Turing machine,
isn't it? I don't> want to argue about requirements for
intentionality or whether TMs are> capable of intelligence,
but this TM is not part of the negotiations> regarding our
conventions.No intelligence required.My TM just adds all the
numberstogether (plus 1 for each addition).A TM is too stupid
to know thatthe sum is supposed to be infinity.> (2) Every
natural number has a> representation in our convention, so the
representation cannot be> non-redundantly extended.I just
described such a TM.No tape can contain every representation
of a natural number.There will always be a bigger
representation.Russell- 2 many 2 count===
===
Subject:
: Re: x^2 +
y^4 = z^4
<fab2a28f.0402180749.38b7c574@posting.google.com>===
===
Subject:
:
Re: x^2 + y^4 = z^4 > the equation x^2 + y^4 = z^4 has no
positive integer solutions. > Is the proof ... short enough
for some kind soul to post it >An elementary proof is based on
descent. That is, assume a >solution exists in positive x, y,
z, and show that another one can be >found with smaller z.Are
you sure?That's true of x^4 + y^4 = z^2 where infinite descent
is used
athttp://www.math.toronto.edu/mathnet/plain/questionCorner/
fermat4.htmlHowever I see no way to modify the method used
there to this problem.----===
===
Subject:
: Re: JSH: Apology to
Ramsay, why I postNora Baron...> You denied and ignored the
theory. You refused to learn> it. The arithmetic, however, you
were ultimately able to handle and> could not deny. That is the
level at which you are able to operate.Bear in mind that Harris
lives through his computer. E.g. for him afunction is something
done in Java or some other computer language. Harrishas no
interest in mathematics, for if he did he would know by now
what analgebraic integer is, for example.> ... Someone said it
previously: you have to be force-fed the truth,> little bits at
a time ...I have a different theory. Harris is willing to deny
anything indefinitely,but he will admit that this-or-that is
true when he fears that his audienceis losing interest in his
latest error or fiction. He needs an audience, buthe has no
interest in mathematical facts. This unusual apology of his
ismotivated only by the unusually heavy pounding that he has
been takinglately. He needs that audience. He has nothing
else.>...> Perhaps tomorrow you will> post something bragging
about your prime counting theorem and your> partial difference
equation, trying to draw attention away from this> major
failure...Probably.> But I am fairly certain you will be back
on this topic as well...I have no doubt of it at all. He will
get away from it for a while, but itwill gall him day and
night, and when he reckons his latest tantrums areforgotten,
he will start the same thing all over again, just as you
say.-- Larry Profiler Extraordinaire Hammick===
===
Subject:
: Re:
the anticlassicalist }{ i: linguistic negation
<40325A2C.21936CE9@hate.spam.net>
<103513r3gafocff@corp.supernews.com>
<40328829.1E5C341D@hate.spam.net>
<1035esdt30h4d38@corp.supernews.com>
<m1g5UWHQ2zMAFwsg@baesystems.com>
<c11u1n$rcl$8@hercules.btinternet.com>In message
<c11u1n$rcl$8@hercules.btinternet.com>, Franz Heymann > In
message <1035esdt30h4d38@corp.supernews.com>, galathaea>Of
course, you also snipped all the references to math and
physics,> Did I blink? I haven't seen any physics yet.>You
should have realised that anything full of -ism and -ist which
is spewed>to sci.lang and sci.logic is not going to contain any
physics.(Nor language; I can't speak for sci.logic ;-) I'm well
aware of that. What I was wondering was whether the _OP_
realises that there's a difference between these isms and
physics, and what it consists of.-- Richard Herring===
===
Subject:
:
Polynomoa; solutions to Pell eqnThe Pell equation and the
obvious cubic generalisationX^2 - DY^2 = 1 andX^3 + DY^3
+(D^2)Z^3 - 3DXYZ = 1have simple polynomial solutions(n,1) for
D = n^2 1(n^2, n, 1) for D = n^31In the square case, you can
use continuedfractions to expand expression of the form,
say,SQRT(an^2 + bn + c), but in the cubic caseno such option
is available.Simple cases can be guessedfor exampleD = n^3  3X
= n^6  3n^3  1Y = n^5 2n^2X = n^4  nD = n^3 +2, n evenX =
(9n^6)/4 + (9n^3)/2 +1Y = (9n^5)/4 + 3n^2Z = 3n(3n^3 +2)/4I
was wondering if there is anysystematic approach to
obtainingsolutions ?===
===
Subject:
: Axioms defining a finite
fieldLet (F, +, *) be a finite set with two operations and
constants 0, 1such that the following rules hold:(1) a + (b +
c) = (a + b) + c(2) a + 0 = a(3) for every a there's a b so
that a + b = 0(4) a*(b*c) = (a*b)*c(5) a*1 = a(6) 1 is
distinct from 0(7) a*(b + c) = a*b + a*c(8) (a + b)*c = a*c +
b*c(9) a*b = 0 => a=0 or b=0Show that F is a field. Can one of
the rules be omitted so that Fstill has to be a field? Posted
Via Usenet.com Premium Usenet Newsgroup
Services------------------------------------------------------
---- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY
**----------------------------------------------------------
http://www.usenet.com===
===
Subject:
: Re: JSH: Making it personal>
Dik Winter keeps up webpages on some of my OLD failed attempts
at> proving Fermat's Last Theorem.I'm sure we can find some NEW
failed attempts of yours if you want.You really ought to take
care on punctuating things some times.===
===
Subject:
: Re: the
anticlassicalist }{ vi: into the quantum> Before you lies the
Void.To toss a chuck of reality into your cranial cave...> I
want to address ontology.then get real. ;-)===
===
Subject:
: Re:
Axioms defining a finite fieldwhy not try and do your own
homework?===
===
Subject:
: Re: No Set Contains Every Computable
Natural <keSdnWEKapvdv63dRVn-uQ@comcast.com>
<LhXXb.13180$gP1.7891@newssvr29.news.prodigy.com>
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<87n07fo9qr.fsf@phiwumbda.org>
<A-mdnYi_UPeU4qnd4p2dnA@comcast.com> Discussion, linux)> Given
any set, S, of representations of natural numbers,> I can write
a TM that will find a representation of a natural number> that
is not in S. (No, this TM does not halt after a finite number>
of steps.)> This is, of course, simply false, false, false.> If
you give the set of all representations of natural numbers,
then> your TM will not find any representation of a natural
number not in> your set.> You're speaking nonsense again.> Of
course, we must be careful here and fix a convention for when
a> tape represents a set of natural numbers. Let's say that a
tape> represents a set of natural numbers if it consists of
blocks of ones> separated by a single space.> We must also be
careful to require that your TM has the property that,> for
each square x of the tape, there exists a steps t such that
for all> steps s after t, the symbol on the square x at time s
is the same as> the symbol on x at t. After all, a TM which
repeatedly changes the> 0th square to 1 and blank and 1 and
blank cannot be said to produce> any tape at all.> I have
given this proof several times.Yes, and you've been simply
wrong several times.> My TM has the property you describe.>
Now, on an input tape containing 1, ,1,1, ,1,1,1, ,1,1,1,1,
,...,> How about b1b11b111b1111...> there is simply no
possibility that your TM, even after an infinite> number of
steps, creates a tape with a representation of some natural>
number not on my input tape. Every natural number is
represented on> my tape. None is missing. Your tape either
returns a set of numbers> already on my tape or it returns no
set of numbers at all.> This TM will find a representation not
on your tape:> It is a three state machine and I can provide a
state> transition table if you like.> 1) Find a blank> 2) Find
a second blank> 3) Backup and write a 1 on the previous blank>
repeat steps (1) through (3)> This TM will produce a tape that
contains exactly one blank.> The contiguous string of 1's
preceding this blank will be> a representation not on your
tape.It will produce no such tape. It will produce a tape
consisting ofall 1's. This is obvious.Suppose that it contains
a blank. Then that blank must occur at somesquare on the tape,
say square n. It is trivial to see that, by thenth iteration
of your three steps, square n is no longer blank.(After the
first iteration, square 1 is not blank, square two wasnever
blank, square three is not blank after the second iteration
andhence is also not blank after the third, and so on.)Your
proof of this claim is simply another confusion.> No, no, no.
A TM cannot extend the representation of natural numbers> for
two reasons: (1) It's a ing Turing machine, isn't it? I
don't> want to argue about requirements for intentionality or
whether TMs are> capable of intelligence, but this TM is not
part of the negotiations> regarding our conventions.> No
intelligence required.> My TM just adds all the numbers>
together (plus 1 for each addition).> A TM is too stupid to
know that> the sum is supposed to be infinity.If your tape
contains a single contiguous block of an infinite numberof
1's, then your tape does not contain a set of natural
numbers.Nothing can change that fact *except* changing what
our conventions oftranslating tapes to subsets of N and back
are. Your machine isinvoked only after the conventions are set
and cannot change anyconventions.It is obvious that your
machine does not produce a (tape representinga) set of natural
numbers even though at each finite step the tape themachine is
working on represents a set of natural numbers.> (2) Every
natural number has a> representation in our convention, so the
representation cannot be> non-redundantly extended.> I just
described such a TM.> No tape can contain every representation
of a natural number.> There will always be a bigger
representation.You have not extended the representation of N.
You haven't evendescribed correctly what tape your machine
produces. If you coulddescribe what your machine produces and
if you were capable ofunderstanding, then you would realize
that your machine does notproduce a tape representing a set of
natural numbers.Of course, you have been making the same
rudimentary and embarrassingmistakes regarding transfinite
sets for literally years. I don'texpect you to learn a damned
thing from this exchange either. Toobad.-- However, you
presuppose that certain numbers *are* prime ideals,... when in
fact ...* they are not... (Maybe I should look up 'primeideals'
but the effort doesn't seem to be worth it. I assume someposter
will get excited ... if I messed up.) --James Harris===
===
Subject:
:
Re: No Set Contains Every Computable Natural
<LhXXb.13180$gP1.7891@newssvr29.news.prodigy.com>
<odydnXiTFf4ZQqzdRVn-ig@comcast.com>
<dzmYb.4044$d34.673767@news20.bellglobal.com>
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<3_OdnTxoJNyqVK_dRVn-hA@comcast.com>
<c0ulem023pj@drn.newsguy.com>
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<c0uql702kub@drn.newsguy.com>
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<aa503d8.0402180923.4fcd297c@posting.google.com>
<UPudnXuOUpAmVa7dRVn-uQ@comcast.com>
<c10mt202f0m@drn.newsguy.com> Discussion, linux)> If you want
to be more general than the usual treatment,> you can do the
following:> Let A be some finite alphabet (containing 0 and 1,
at least),> and let A^omega be the set of infinite sequences of
elements of A.> Every such infinite sequence is a possible
state> for the infinite tape of a Turing
machine.[...]regarding the same issue (N is recursive as a
subset of N, but the setof tapes consisting of representations
of natural numbers is notrecursive as a subset of all possible
input tapes). It would be nicehere. Otherwise (?), he's simply
diddling and wasting the time of everyoneinvolved.-- So, at
this time, I'd like to assure you that I am not interested
inmaking sure mathematicians worldwide get fired. I've
rethought mydesire to go to Congress and try to get funding
for mathematicianscut. -- James Harris is a reasonable man.
Whew! ===
===
Subject:
: Re: combitorics> I'm not sure if this is a
good group to post this.> I have a combinatorics problem.
well, a puzzle really. I'm not sure> what kind of problem this
is. so I would love to get some feedbacks> from the group. ( as
to the nature of the problem as well as how to> solve it)>
Let's say you have a little sister named Torrey. And she has
quite a> quite a number of friends in kindergarten. let's say
her friendship> goes as follows.> Torrey loves Matt on some
days, but Torrey hate Matt on other days.> Torrey also loves
Paul on some days, but Torrey hates Paul on other> days.>
Torrey also loves Sandy on some days, but she never hates
Sandy.> Sandy hates Torrey on some days, but loves her on
other days> Matt loves Sandy on some some days, but hates her
on other other days> Paul Loves Matt on some days, but Paul
hates other days.> paul hates Sandy on some days. > to make it
a bit simpler...> Torrey loves and hates Matt> Torrey loves and
hates Paul> Torrey loves Sandy > Sandy loves and hates Torrey>
Matt loves and hates Matt> Paul loves and hates Matt> Paul
hates Sandy> how can I figure out how many possible
combinations there are that> desribes the love and hate
relationship between Torrey, Matt, Paul,> and Sandy?> I have
32. but I'm not sure if i'm right. and i'm doing it brute
force> way.> any helpful thoughts and comments are
appreciated. > thank you all. > seanyou're right. 32 = 2^5.
There are two possibilities: either torreyloves matt or hates
him. In either case there are two morepossibilities: either
torrey loves Paul or hates him. After thesecases we have 2 * 2
= 4 possibilities:(love, love), (love, hate), (hate, love),
(hate, hate). Imagine a 2x2square with each of these in them.
Clearly if you multiply the numberof possibilities for the
first thing (2) by the number ofpossibilities for the second
thing (2), you get the total number ofpossibilities. Now there
are 5 choices, so the total is 2^5 = 2 * 2 *2 * 2 * 2 =
32.===
===
Subject:
: Re: question about periodic function> I am
having trouble empirically determining the period of the
followingfunction:y = sin(2*pi*.018*t) + cos(2*pi*.02*t)It is
a situation of superposition of two waves with a
slightlydiffering frequencies (.018,.02). One can hear beats
in a tuning forkmathematical sense it is strictly not a
periodic function with periodT as the definition f(x+T)=f(x)
is not satisfied. But the followingapproximation is accepted
in electronics engineering wave formanalysis and acoustics.Let
2 pi f t =w t be the angular argument. sin(w1 t) + cos (w2 t) =
cos(pi/2- w1 t) + cos (w2 t) = 2 cos(pi/4+(w2-w1)t) cos(pi/4-w
t) where w's w1~w2~w areapproximately equal.w2-w1 is neglected
in comparision to pi/4, making it sqrt(2)cos(pi/4-w t).So it
has maximum value sqrt(2) at wt=pi/4 and regular(high
frequency) time period 2 pi/w and beating time period
2pi/(w1-w2) appearing as the envelope in the compound function
graph.Works out to 1/.019 and 1/.002 or 52.6 and 500 seconds of
regular(hf)and beat periods respectively.===
===
Subject:
: Re: the
anticlassicalist }{ vi: into the quantumI want to address
ontology.> then get real. ;-)There _is_ a problem there.
Should it beMr Ontology, Mrs Ontology...?I was about to
suggest Comrade Ontologybut... ah, the English speaker's lot
isnot a happy one! If only we spoke Japanese: Ontorogii-san.
Done!(Why Ontorogii and not Ontorojii?Dunno. Sounds
better).===
===
Subject:
: Re: No Set Contains Every Computable
Natural (was Church-Turing compared to Zuse-Fredkin thesis)>
<muse> I'm remembering Wegner's interaction machines. This
may be a completelift> of Wegner, but here's an idea. Define a
so-called 'stream' TM that canread> a possibly infinite input
stream and produce a possibly infinite output> stream. The TMs
being discussed would then befinite-input/infinite-output>
stream TMs. So instead of 'non-halting', you would say
'streaming'.> However, a stream TM would have both an internal
tape, and a write-only> output tape. Any TM that can (*)-output
its stream to any finite> specified length, in finite time, can
be modified to use a seperate tape> for output in an obvious
way.> To be a stream TM a machine must have one or both of :
infinite output or> infinite input. Stream TMs can be either
infinite or finite state. TheTM> discussed would have to be
infinite state. Any finite state infinite-out> stream TM has
an output that eventually repeats forever, therefore its>
output set must be countable; each element representable by a
rational.As> a generalization of Turing's rule (*), I propose
: (**) for every finite> prefix O of the output, there exists
a finite prefix I of the input, such> that O appears on the
output within a finite number of steps after I is> read.> In
the case of a stream language L, (and I definitely recall
reading about> this in a Wegner paper), we can define L: input
-> output as L: R -> R, at> least for some inf-in/out infinite
state TMs. But where Wegner gets it> wrong is to imply that
since we seem to be mapping reals to reals,> interaction
machines are superTuring. What he fails to realize is that>
although the cardinality of the languages recognized by these
machines has> gone from countable to uncountable, we have not
upgraded our definition> stating a computation must take a
finite number of steps. (Note, it makes> sense to replace
'steps' with 'time', since steps implies a particularmodel> of
computation). Maybe call this notion limit computable, or
stream> computable. For example, although it may be able to
stream-compute pi, a> stream TM is not superTuring unless it
could output pi in finite time.> Languages would probably end
up being defined in terms of the output set> instead of the
input set, since one can no longer accept or reject aninput>
for any inf-in stream TM. This would fit perfectly in terms of
TMs that> stream compute numbers like pi, since the important
stuff is in the output> not the input.> </muse>Note, mapping
reals to reals assumes the input can be an
uncomputablesequence. Without this, even
infinite-input/infinite-state stream TMs couldonly output a
countable set, since the computable reals are countable. Ifthe
input comes from nature, I assume most would say this was
possible,assuming an infinitely existing universe. But any
harnessable physicalprocess that could generate this input
could be considered to be, at leastpart of, a physically
realizable superTuring computer.This also seems to imply that,
regardless of physical reality, no stream TMin any closed
network of stream TMs (each allowed any finite number
ofseparate inputs and outputs) can ever receive or output an
uncomputablenumber. This can be seen from the fact that any
such system can be seriallysimulated by a standard TM.l8r,
Mike N. Christoff===
===
Subject:
: Re: JSH: Apology to Ramsay, why I
postsays...>It's simpler to just post to all the newsgroups
that I posted before>an apology for questioning Keith Ramsay's
honesty. It seems he did in>fact post a non-unit algebraic
integer factor of (1+sqrt(-167))/2 and>7, so I was wrong.>So I
apologize to Keith Ramsay for questioning his honesty here.Take
the next step in the learning feedback loop: remember this
incident andnext time someone disagrees with you don't
automatically accuse them of lying.===
===
Subject:
: Re: Klingon,
anti-universals etc. (Re: the anticlassicalist }{ i:
linguistic negation)Okrand says he violated a few human
language universals in inventingKlingon, since of course it
wasn't meant to be a human language.> I've got the Klingon
dictionary (which is actually a handbook),> I've got the
tapes, I've got the later Klingon Berlitz Guide> (not its true
title, it's the Klingon dictionary again,> adapted for galactic
hitchhikers, can't remember its real> title), I've been through
the lot, I haven't come across> anything even remotely weird.
If you want something really> weird, try Lojban instead.> (No
apologies for cross-posting to alt.phil, sci.m, sci.l,> sci.p,
someone out there started this , not me)Yp, I've got the
dictionary, too, and my recollection from reading itquite a
few years ago is that there was nothing glaringly weird
aboutit. I was just quoting something Okrand said in an
interview. Someoneonce told me that he had based it on some
indigenous language ofnorthern California, which would not be
implausible, since many graduatestudents at Berkeley in those
days worked on those languages. But I havenot heard anything
more specific than that, so it can't be too obvious,even if he
got some ideas from there.Ross Clark===
===
Subject:
: Re: the
anticlassicalist }{ i: linguistic negation> Klingon was
specifically created to be the worst language possible by>
folks who knew linguistics. It was enthusiastically embraced
by the> mob and it is as good as Korean or Chinese for
transferring content.Okrand says he violated a few human
language universals in inventingKlingon, since of course it
wasn't meant to be a human language. Nothingabout worst
language possible. Studies of how Klingon-users actuallyspeak
might be interesting. Do you know of any? And anyhow what was
thismeant to prove, even if it were all true? Sure as hell
wouldn't provewhatever you were salivating about in the
previous paragraph anyway.> You can find a significant
community of people who speak Klingon at> the Klingon Language
Institute.> http://www.kli.org/> The archives of the KLI's
email discussion forum might be useful if> you're serious
about studying how Klingon is spoken.>
http://www.kli.org/tlhIngan-Hol/Not me personally, no. But I
can see some interest in looking at whathappens to any conlang
if it is regularly used by a lot of people. Hasanybody tried to
bring up a child as a native speaker? Now that would
beinteresting -- though they probably wouldn't want to
advertise the factto the local child welfare authorities. If
Okrand really did build somecounter-universal features into
it, how a child would deal with it as afirst language would be
really fun to watch.Ross Clark> As for worst language possible,
that would be INTERCAL. Klingon is> actually a very easy
language. Compared to natural languages, it's> essentially a
toy, but it's a very powerful and interesting toy.===
===
Subject:
:
Re: Klingon, anti-universals etc. (Re: the anticlassicalist }{
i: linguistic negation)Okrand says he violated a few human
language universals in inventingKlingon, since of course it
wasn't meant to be a human language. > I've got the Klingon
dictionary (which is actually a handbook),> I've got the
tapes, I've got the later Klingon Berlitz Guide> (not its true
title, it's the Klingon dictionary again,> adapted for galactic
hitchhikers, can't remember its real>
title),...(_Conversational Klingon_ and _Klingon for the
Galactic Traveler_)> I've been through the lot, I haven't come
across > anything even remotely weird.The standard example of
weirdness is the division of color words. Nohuman language
ever splits things the way Klingon does, groupingred/orange in
one word and yellow/green/blue in another.There's also some
major weirdness in the inventory of sounds. There'sa standard
English /t/ but no corresponding /d/; the retroflex voicedstop
that takes its place has no unvoiced counterpart. The weird
partis not in what sounds are used, but rather in what sounds
are missing.===
===
Subject:
: Re: JSH: Making it personal>I'm
concerned about all the histrionics that has darkened even my
own>posts as it's all too personal at this point._You're_
concerned about things being too personal?People spend weeks
trying to point out a simple counterexample tothe things
you've been saying. You accuse them of lying when they'renot
(at worst they're making an honest mistake, and in fact
they'renot even doing that, they were _right_.) You state that
nobody herehas any regard for the Truth. And now you complain
about thingsbeing too personal. ing asshole.>Right now I
*despise* posters like Dik Winter, Nora Baron, and
David>Ullrich, which necessarily involves some fear: they're
potent at>fighting for their positions.Wow. It seems like
you've _already_ forgotten about yesterday, whenyou finally
conceded that their position was actually _correct_.>They also
just so happen to ultimately be wrong.Huh? Wrong about what,
exactly? ing lying slanderous asshole.>Dik Winter keeps up
webpages on some of my OLD failed attempts at>proving Fermat's
Last Theorem.>That's just bad behavior.Because you find it
embarrassing? Would you prefer web pages onyour NEW
failures?HINT: Your _own_ page is equally embarrasing, equally
full ofnonsense, as those old pages. The only difference is
that you'renot aware of it, because you're a ing
idiot.(Getting the math wrong over and over doesn't make you a
ingidiot, a lot of people make a lot of mistakes. But
insisting _every_time that you're right, and _every_
professional mathematicianon the planet is wrong, forgetting
about the _many_ times inthe past they've turned out to be
right, makes you a ingidiot.)>Trotting out someone else's
mistakes made years ago, and pushing them>on your own personal
webpages is not something that's defensible.>It's clearly
making a statement.That's correct. Here's the statement:
You're a fool. That's clear.>Dik Winter invites my contempt,
and unfortunately, I have to fear the>effectiveness of such
tactics.>It's sad really.>I'm going to see about dropping that
fear as I think it gets in the>way of research, but right now I
do despise these people because>they're so contemptible, and so
successful in screwing up the>dialogue.What research? You've
never actually got anything right. Screwingup what dialog?
Over and over people point out you're wrong about something,
and then a dialog ensues, consisting of youinsisting you're
right and ignoring the explanations that you'rewrong. Every
time you _eventually_ agree that you're wrong.But regardless
of how many times it happens that it turns outthe people
you've accused of _lying_ about the math wereactually right,
your retraction is always no big deal. Thislast time you
included an apology for questioning Keith'shonesty, just for
show, but you omitted the apology forquestioning the honesty
of all those other people, whoyou now _know_ were simply
telling the truth.And no matter how many times it happens,
after youfinally admit you were wrong you _still_ have
complaintsabout the fact that people have been simply
explainingwhy you were wrong, during the months that it
tookyou to catch on.Hence the phrase ing asshole.>James
Harris************************===
===
Subject:
: Re: Axioms defining
a finite field===
===
Subject:
: Axioms defining a finite field >Let
(F, +, *) be a finite set with two operations >and constants
0, 1 such that the following rules hold: >(1) a + (b + c) = (a
+ b) + c >(2) a + 0 = a >(3) for every a there's a b so that a
+ b = 0 >(4) a*(b*c) = (a*b)*c >(5) a*1 = a >(6) 1 is distinct
from 0 >(7) a*(b + c) = a*b + a*c >(8) (a + b)*c = a*c + b*c
>(9) a*b = 0 => a=0 or b=0 >Show that F is a field. Can one of
the rules be omitted >so that F still has to be a field?let a
/= 0, b0 = 0, b1 = 1A = { a bj | 0 <= j <= |F| }|A| = |F|
because if a bj = a bk: a(bj - bk) = 0; bj - bk = 0; bj = bkAs
F is finite, 1 in A which shows a has right multiplicative
inverse a_r.This with a1 = a, shows F0 is a group under *.Did
you forget a+b = b+a, ab = ba?----===
===
Subject:
: Re: Klingon,
anti-universals etc. (Re: the anticlassicalist }{ i:
linguistic negation)> The standard example of weirdness is the
division of color words. No> human language ever splits things
the way Klingon does, grouping> red/orange in one word and
yellow/green/blue in another.Doesn't strike me as weird. I
know a language whichhas no word for orange, no word for blue,
which hasthree words for yellow and green, but does not
differentiatebetween them by colour, but by brightness (or
perhaps bysaturation, I never quite figured it out).> There's
also some major weirdness in the inventory of sounds. There's>
a standard English /t/ but no corresponding /d/; the retroflex
voiced> stop that takes its place has no unvoiced counterpart.
The weird part> is not in what sounds are used, but rather in
what sounds are missing.You have led a linguistically
sheltered life. The Lolovuevue languageof Oba Island has only
voiced, prenasalized stops ([mb], [Ng], [Ngmb]),butno [p], no
[k], no [kp], let alone a [b], a [g], or a [gb]. But it
doeshave a /t/ which is always a straight [t]. And it does not
have a [nd].As for the language of Shark Bay, it has two s's,
but only one ts, andits speakers cannot tell the difference
between s and sh.There is nothing weird in the Klingon
inventory, not major, not minor.===
===
Subject:
: Re: covering
compact set w/squares>[...]>Given a compact set K in the plane
s.t. each pt x is the center of a>square>Q_x, prove that you
can find a subsequence Q_x_i of squares s.t. K is>covered by
the Q_x_i and the sum of the areas of each of the Q_x_i
is>no>more than 4 times the area of the union of the
Q_x_i.Thought about this a little more. Here's a construction
that it seemsto me _may_ do the above, although I haven't
quite proved that itworks:Going to consider the case of finite
K (reducing to this case may ormay not be easy, I think I see
how to do it maybe but haven'tworried much about that, since I
can't _prove_ this case anyway...)Let Q_1 be the largest of the
squares (pick on at random in case ofa tie.) Discard all the
squares with center in Q_1 and let Q_2 bethe largest of the
remaining squares. Discard all the squares withcenter lying in
Q_2 and let Q_3 be the largest of the remaining, etc.Now the
Q_j cover K, since every point of K is the center of oneof the
original squares and that square never got thrown outunless its
center was contained in one of the square that wasalready
selected.And it _seems_ to me that no point (with the possible
exceptionof the center of one of the squares, a set of measure
0) can liein more than 4 of the Q_j. I haven't _proved_ this,
but it seemstrue: Say S_1,...S_5 are five of the Q_j, listed
in order ofincreasing j. Then S_2 is no larger than S_1 and
its centerlies outside S_1. Draw a picture of various ways S_1
andS_2 can look. Now S_3 is smaller, and its center liesoutside
S_1 and S_2... In order for S_4 to intersect theintersection of
the first three S_j it sort of has to lie onthe other side of
S_2 relative to S_1 (the side other thanthe side where S_3 is
located) and now there's no wayto position S_5 thaf allows it
to reach far enough insideto intersect the intersection of the
first four...That's a pretty fuzzy argument, but given the
constraintson the sizes and the location of the centers I
don'tsee how five of the Q_j can
intersect.************************===
===
Subject:
: Re: Axioms
defining a finite field===>
===
Subject:
: Axioms defining a finite
field>Let (F, +, *) be a finite set with two operations>and
constants 0, 1 such that the following rules hold:>(1) a + (b
+ c) = (a + b) + c>(2) a + 0 = a>(3) for every a there's a b
so that a + b = 0>(4) a*(b*c) = (a*b)*c>(5) a*1 = a>(6) 1 is
distinct from 0>(7) a*(b + c) = a*b + a*c>(8) (a + b)*c = a*c
+ b*c>(9) a*b = 0 => a=0 or b=0>Show that F is a field. Can
one of the rules be omitted>so that F still has to be a
field?>let a /= 0, b0 = 0, b1 = 1>A = { a bj | 0 <= j <= |F|
}>|A| = |F| because> if a bj = a bk: a(bj - bk) = 0; bj - bk =
0; bj = bk>As F is finite, 1 in A> which shows a has right
multiplicative inverse a_r.>This with a1 = a, shows F0 is a
group under *.>Did you forget a+b = b+a, ab = ba?I don't know
about the a+b = b+a, but if we add that thenab = ba follows:
it's a theorem that every finite divisionring is a field.
(It's not quite trivial, as I recall. Possiblyit actually is
trivial and just didn't seem trivial to meat the time; that
was an algebra class when I was
anundergraduate...)************************===
===
Subject:
: I got
low score on math test, please advise me and take a lookmy
website states my case and has jpg files of the four pages of
the testplease take a look and advise me or give me
opinionshttp://www.johncho.us===
===
Subject:
: Re: covering compact
set w/squares>[...]>Given a compact set K in the plane s.t.
each pt x is the center of a>square>Q_x, prove that you can
find a subsequence Q_x_i of squares s.t. K is>covered by the
Q_x_i and the sum of the areas of each of the Q_x_i
is>no>more than 4 times the area of the union of the
Q_x_i.>Thought about this a little more. Here's a construction
that it seems>to me _may_ do the above, although I haven't
quite proved that it>works:>Going to consider the case of
finite K (reducing to this case may or>may not be easy, I
think I see how to do it maybe but haven't>worried much about
that, since I can't _prove_ this case anyway...)>Let Q_1 be
the largest of the squares (pick on at random in case of>a
tie.) Discard all the squares with center in Q_1 and let Q_2
be>the largest of the remaining squares. Discard all the
squares with>center lying in Q_2 and let Q_3 be the largest of
the remaining, etc.>Now the Q_j cover K, since every point of K
is the center of one>of the original squares and that square
never got thrown out>unless its center was contained in one of
the square that was>already selected.>And it _seems_ to me that
no point (with the possible exception>of the center of one of
the squares, a set of measure 0) can lie>in more than 4 of the
Q_j. I haven't _proved_ this, but it seems>true: Say S_1,...S_5
are five of the Q_j, listed in order of>increasing j. Then S_2
is no larger than S_1 and its center>lies outside S_1. Draw a
picture of various ways S_1 and>S_2 can look. Now S_3 is
smaller, and its center lies>outside S_1 and S_2... In order
for S_4 to intersect the>intersection of the first three S_j
it sort of has to lie on>the other side of S_2 relative to S_1
(the side other than>the side where S_3 is located) and now
there's no way>to position S_5 thaf allows it to reach far
enough inside>to intersect the intersection of the first
four...>That's a pretty fuzzy argument, but given the
constraints>on the sizes and the location of the centers I
don't>see how five of the Q_j can intersect.Maybe you can
actually prove it, along the following lines:Note that when
talk about, say, a point lying abovea line segment I don't
mean directly above, I meanthe y coordinate of the point is
larger than the ycoordinate of points on the segment.The
center of S_2 is outside S_1; assume the centerof S_2 lies
above the top edge of S_1.Now the center of S_3 cannot lie
above the top edge ofS_2 or the intersection would be empty.
There arevarious cases left. One case would be where thecenter
of S_3 lies to the left of the left edge of S_2.In that case
the center of S_4 cannot lie to theleft of the left edge of
S_3, so it must be to theright of the right edge of S_3 or
below the bottomedge of S_3... you construct a tree of all
thepossibilities and finally determine that there's noplace
for S_5.************************===
===
Subject:
: Re: Advice on
future with MathI'm an upperclass math major that was/is
planning on attempting amasters or phd in math. Here lately,
I've wondered if this is a greatidea.My first question is
this: How much should you study (reading thematerial, working
problems) for, say, an abstract algebra or advancedcalculus
junior/senior level class? I talked to a graduate studentthat
said he studied for a couple of his first year grad classes
atleast 4 hours a day on weekdays and about 8 hours on
Saturday andSunday. Needless to say, I do nothhing like that.
Should I be doingthat or is that just an insane amount of time
to be spending on it?It seems that if you're good at something,
it shouldn't require somuch work.> I might be an anomaly, but I
don't spend much more than 2 hours a day> actually sitting down
with pencil and paper working on math. However,> what I am
doing is almost every minute I am awake and every single>
minute I am asleep, I am thinkingabout math, or one of the
problems I am> working on. Occasionally I'll grab some scratch
paper and jot something> down. Most of the time, not.> Okay,
maybe saying every minute is a little exaggeration, but not
much.> Seriously, though, I spend the best part of my day,
everyday thinking> about math. If I am watching TV or reading
the newspaper, or talking> with someone, in the back of my
head is something like, What is the> solution to the PDE a*u_x
+ b*u_y + c*u_z = 0?> Also, it has been my experience that if
you have to spend insane amounts> of time studying math, then
maybe continuing on is not for you. I am> not trying to
discourage you, and I would hope you at least try for a> year,
but if you are stuggling and spending 8 hours a day in the
books,> you might want to look for something else.Well, I
disagree. In my experience, it is very important to
practice,practice, and practice. This is of course not
particular to math. So, yes, Ispent a large amount of time on
math, many hours a day. It does take hardwork studying math,
unless you're born a genious, but for now I assumeyou're just
like the rest of us :-) You've really got to *like* math. If
itis a pain doing the assignments and reading the books, then
you should havesecond thoughts.Perhaps you should think about
why you're interested in math. To me, math islike a tool, kind
of like a swiss army knife. It can do all kinds of weirdthings;
or rather, *I* can do weird and interesting stuff using this
tool,i.e. math. While I was an undergraduate, I had a
different attitude. It wasan intellectual challenge: I was
*determined* to prove (to myself, if notanyone else) that I
could solve the problems. So basically, I just keptgoing and
going and going, until I had solved a problem.In high school,
math was very easy for me, and the only homework I did wasthe
written assignments. When I reached graduate level, it
suddenly becamequite difficult. Sometimes I had to think about
a proof for days or weeks,before I found the right path, or
even understood the solution. It helped meimmensely that I was
studying with some likeminded people.> Also, there are many
different branches of math. Personally, I am of> the
discrete/combinatorics and numerical methods persuasions.
Analysis,> topology and PDE, I only took those classes becaues
they were required.> And, I struggled through them.Good point,
I completely agree. No-one is good at everything, so you
mustexpect that some branches give you the kick and interest
you, while otherseither bore you or are too difficult to grasp
(at the moment).At times I have difficulty understanding things
that, after findingout where I went wrong, seem pretty simple.
I assume others don'thave this problem because normally they
can respond right away withwhat some math concept meant or,
less often, help me where I am havinga problem understanding
part of a proof. Is it normal to not follow aproof or just a
sign that I'm just weak mathematically (we're againtalking
junior/senior level advanced calc/abstract algebra here)?> If
after finding out where you went wrong things are still
difficult,> you are, well, screwed. The right answers always
look easy. Then you> think, Who the heck thought of doing it
that way? However, that is> not to say you shouldn't be able
to follow a proof. I read many papers> and I can't always
follow the proofs givne in them. Often, they skip> steps or
say somethign is obvious when it isn't (unless you know som>
obscure theorem). But, if the proof is in your text, you
should, for> the most part, be able to follow it.Let me just
repeat: Practice, practice, and practice. Here's a
suggestion.Dig out some textbooks that you used a couple of
years ago, and read themagain. You should expect two benefits:
The first is repetition, the secondis a realisation that things
aren't as difficult now as they were then.I have a bad tendency
to be hot/cold when it comes to math. It seemslike I'll have
the highest/near highest score on an exam and thenreally screw
up on the next exam, getting something in the bottom halfof the
grades. I wish that I could attribute it to something likewhen
I first started school: I would do no work until after I
screwedup on the first exam, and then work harder after that
and do quitewell. Now it seems that I'll do well on the first
exam and then screwup on the next one. I really don't think
that I will have studiedmuch less for the second one but I
guess it could be a possibility.That was my first thought.
However, begin hot/cold should not frighten you;we can all
have out down periods. Don't look at the results, but
considerinstead what it's like studying. Is it fun or is it
tedious? Again, don'tfocus on isolated bad days, we all have
them. Think about the big picture.Look back at last semester,
what was it like taking the math courses?When I do bad on the
other exam, I begin to wonder if I just gotlucky on the
problems that were given on the good exam and would havedone
lousy if a few different problems were picked.Argh! You must
believe in yourself. Whenever you do good in an exam,
giveyourself a pat on the shoulder. You've deserved it!> Okay,
those are my thoughts. Take them and do what you will with
them.> But, I tyhink if you reall ylike math, and are not a
burnt out senior> who is just looking forward to graduation,
you should at least try a> semester working towards a master
degree. If you find out it's not for> you, well, you learned
and you didn't really waste any of your life.> But, if you
don't try, you will always be kicking yourself.I completely
agree.===
===
Subject:
: [OT]Re: No Set Contains Every Computable
Natural> --Come on people!!! The US just blew up a lot of
people in Iraq, don't> you realize that a person with my
exposure might just end up dead, by> mysterious
circumstances?> --James Harris, on the dangers of proving
Fermat's last theoremThese quotes are classic :) Note that in
the 'shaquille o'niel in your backyard' quote, its funnier if
you put his actual intention in. ie: not an'internet
personality', but 'a gifted thinker who's better at math
thanyou'. :)l8r, Mike N. Christoff===
===
Subject:
: Re: question
about periodic functioncontinued..However, the bounds are +/-
2 as it is product of two cosinefunctions, and not sqrt(2) as
erroneously stated above. Theapproximation procedure is valid
for time/frequency purpose only, notamplitude!===
===
Subject:
: Re:
How big can a manifold be?> well its emmbeddable in R^3 its
obvious of cardinality c, no?But the embedding theorem assumes
it is second-countable(or, equivalently, paracompact). What
ifwe omit that assumption?--
http://www.math.ohio-state.edu/~edgar/===
===
Subject:
: Pascals
TriangleThere was an interesting puzzle in the Sunday Times in
the UK recently thatset me thinking about the issue that the
puzzle raised.The idea is that of a set of points arranged in
the familiar Pascal'striangle format (rows 1, 2, 3, ... etc.
containing 1, 2, 3, ... points in atriangular format) after
which the issue is that of the total number ofequilateral
triangles that can be found in this set of points.This itself
is not hard but the subsequent question seems more difficult
-for such an arrangement with N rows, what is the minimum
number of pointsthat need to be blocked in order to reduce
this total number of equilateraltriangles to zero?I wondered
if anyone here might have come across this or have some
insightsabout a general solution. Gladman===
===
Subject:
: Re:
Pascals Triangle> There was an interesting puzzle in the
Sunday Times in the UK recently that> set me thinking about
the issue that the puzzle raised.> The idea is that of a set
of points arranged in the familiar Pascal's> triangle format
(rows 1, 2, 3, ... etc. containing 1, 2, 3, ... points in a>
triangular format) after which the issue is that of the total
number of> equilateral triangles that can be found in this set
of points.Pascal's triangle is 1 1 1 1 2 1 1 3 3 11 4 6 4
1...and does not contain points, as such; it contains
numbers.Row 2 does not contian a 2, but sums to 2. Row 3 does
not contain a 3 orsum to 3. You've completely lost me.Phil--
1st bug in MS win2k source code found after 20 minutes:
scanline.cpp2nd and 3rd bug found after 10 more minutes:
gethost.cBoth nonxploitable. (The 2nd/3rd ones might be,
depending on the CRTL)===
===
Subject:
: Re: No Set Contains Every
Computable Natural (was Church-Turing compared to Zuse-Fredkin
thesis)> Note, mapping reals to reals assumes the input can be
an uncomputable> sequence. Without this, even
infinite-input/infinite-state stream TMscould> only output a
countable set, since the computable reals are
countable.Allowing uncomputable inputs, would be akin to
standard TMs withuncomputable oracles. Basically, one has to
wonder about, even thetheoretical benefit, of defining the
input set as being uncountable, giventhat an uncountable
number of those inputs will be both uncomputable, and(more
importantly in terms of an ability to do algorithmic and
output setanalysis) undefinable as well.l8r, Mike N.
Christoff===
===
Subject:
: Re: No Set Contains Every Computable
Natural (was Church-Turing compared to Zuse-Fredkin
thesis)messageNote, mapping reals to reals assumes the input
can be an uncomputablesequence. Without this, even
infinite-input/infinite-state stream TMs> couldonly output a
countable set, since the computable reals are countable.>
Allowing uncomputable inputs, would be akin to standard TMs
with> uncomputable oracles. Basically, one has to wonder
about, even the> theoretical benefit, of defining the input
set as being uncountable, given> that an uncountable number of
those inputs will be both uncomputable, and> (more importantly
in terms of an ability to do algorithmic and output set>
analysis) undefinable as well.I realize this is gross
over-posting, but I just had to add this quote fromMathworld:
In fact, all theorems of calculus remain true if the field of
real numbersis replaced by the field of definable numbers,
sequences are replaced bydefinable sequences, sets are
replaced by definable sets and functions bydefinable
functions.Quite interesting.l8r, Mike N. Christoff===
===
Subject:
:
Re: the anticlassicalist }{ i: linguistic negation: >You
should have realised that anything full of -ism and -ist which
isspewed: >to sci.lang and sci.logic is not going to contain
any physics.: >: (Nor language; I can't speak for sci.logic
;-): I'm well aware of that. What I was wondering was whether
the _OP_: realises that there's a difference between these
isms and physics, and: what it consists of.Installment vi
details the properties of the closed linear subspace latticeof
a Hilbert space, detailing the relationship with observables
andprediction in quantum systems. This builds off of the
mathematical theoryYou do understand the physical content, do
you not?--
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-galathaea:
prankster, fablist, magician, liar===
===
Subject:
: Re: Graph Theory
Textbook> I'm an undergraduate, and this summer I will be
participating in an> REU for discrete math and combinatorics.
I am looking for a good> graph theory textbook to learn the
basics from. I have seen the books> that Dover (which of
course are very cheap) has to offer on the> subject, but
unfortunately they seem too elementary. Is the Springer> GTM
Graph Theory by Reinhard Diestel any good? Any suggestions
are> highly welcomed.> Jack[Obvious disclaimer: There are many
good books, I've listed mypersonal favourites].I like Graphs
and Digraphs by Chartrand and Lesniak (3rd edition)very
much.Other good introductory books, in addition to those Jim
and Tim havealready mentioned, are Graph Theory by Harary (an
old classic butstill amazingly relevant and accessible),
Introductory Graph Theoryby Chartrand and the more recent
Graph Theory by Russell Merris andGraph Theory by Douglas
West. Also, I haven't yet read much ofDiestel's book but what
I have is excellent.HTH,Felix.===
===
Subject:
: Re: Sets That
Resemble Derivatives Somewhat by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
i1JDEKP11801;>Define:>1) (A-->B) = (AB')' = A' U B>for any two
sets A, B, where ' is complement and Adjacent letters>are
intersected. This is the set analog of logical a-->b for a,>b
propositions, the latter defined as ~(a ^ ~b) = ~a V b
withBefore replying to the other comments on my previous
posting, whichI prefer to do tomorrow morning (night debates
aren't my favoriteway of going to sleep at age 65, whether
positive or negative), Iwould like to point out some things
about complements.1) A U A' = UniverseThis says intuitively
that a set (A) and/or its complement (thepart of the Universe
outside the set) comprises the Universe. Morespecifically, it
comprises the Mathematical Universe, not necessar-ily the
Physical Universe.If we translated the idea behind this into
Real or Complex Analysis,or even Tensor Analysis and so on, we
would get something like:1') A variable and/or a function and
its change constitute Universeof study or Knowledge about the
variable/function.This is what Garrett Birkhoff of Harvard
meant by saying thatDifferential Equations contain
causation/causality. Of course, incausation/causality the
change is usually with respect to time, buteven in spatial
change (partial derivatives in general) it takestime in a
sense to scan different spatial regions - even on alight wave,
where time is supposed to be instantaneous, sometemporal notion
of the order of attention to different placescan be introduced,
since for one thing simultaneous attention tomany different
places contains some paradoxes depending on howintense the
attention is. I'll mention one more aspect:2) (A')' = AThis
doesn't correspond to differentiation, but I don't think itis
the most critical feature of differentiation. It is more like
d^2 in exterior algebra. Intuitively speaking, when A is aset
and ' is complement, we could translate this as:2') The change
in a change of A is the change back to A.We would, I admit,
sacrifice second order derivatives and higherin this scenario,
but we already know that higher order equationscan often be
expressed as systems of first order equations. Ithink that the
advantages outweigh the orders so to speak, andwe may
eventually find a generalization of the usual derivativeorder
in some other ideas. I especially like the fact that
theRiccati Differential Equation (with of course its Algebraic
version)which underlies Growthxpansion-Contraction processes
and events(e.g., in biology - see the Logistic and Simple
Positive or Neg-ative Exponential Differential Equations which
are special casesof the Riccati Equation, and the
generalization to a Bernoulliequation) is first order.
Quantitative Biology is a new and veryvaluable branch of
Arxiv.org. I'll try to discuss the differencesbetween
Curvilinear One-Direction-At-A-Time motion and
SimultaneousMany-Directions-At-A-Time
Growthxpansion-Contraction Motion(including radiation,
expansion of the Physical Universe as awhole, etc. - possibly
with different rates and amounts in different directions,
relating to (optimal) control also) later.Osher
Doctorow===
===
Subject:
: Re: errors in an argument by
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id i1JDELd11866;> As far as I know the
first flying creatures they talk about are > dinosaurs, no
less. And quite big ones as well.>That's a comment on your
knowledge. Some dinosaurs were small; a quick>search turned up
some with a mass below 10 kg. And all the references I>found
talk about small dinosaurs as the starting point for the
evolution>of wings.>-- >Daniel W.
Johnson>panoptes@iquest.net><a
href=http://members.iquest.net/~panoptes/>http://
members.iquest.net/~panoptes/</a>039 53 36 N / 086 11 55 WI
admit that my knowledge of the research of small dinosaurs
isn't very deep, so you may be right, but that keeps my
general point intact. The way it seems to me (and I'm not
talking as an expert in aerodynamics either) is that even for
extremely small and light creatures, the ability to develop a
functional limb that will allow it even the smallest
maneuverability by mutation benefit is unlikely. Remember that
it has to be aerodynamic to some extent, and to be evolutionary
beneficial. It's very hard for me to imagine how such a thing
would look like.===
===
Subject:
: Re: Parity Check Matrix of a
Systematic Linear Block Code windows-nt)> The generator matrix
of a systematic linear block code has the> form G = [Ik : P].
How can it be shown that the parity check> matrix is of the
form H = [-P^T : In-k]?> With great ease.> If a vector (a : b)
is supposed orthogonal to the code generated> by G then (a :
b)G^t = 0 so a + b P^t = 0 or a = -b P^t etc.> Are you in the
IEEE?Yes. Why do you ask?--
http://home.earthlink.net/~yatescr===
===
Subject:
: Are the
derivatives of abs[(x-a)^3] different for x>a and x<a ?given
abs[(x-a)^3], where abs[ ] means taking the absolute value,
anda is a constant, are the first, second, third derivatives
(withrespect to x) different for x>a and x<a ?I forgot how to
write down the derivative of an absolute function in aformal
way, I mean, not using if else, but using something
likesign(x-a).I think the first derivative would look
like:3(x-a)^2 * sign(x-a)But then I don't know how to proceed
to the second derivative.===
===
Subject:
: Re: Advice on future with
Math> Often, they skip > steps or say somethign is obvious when
it isn't (unless you know som > obscure theorem). There is an
old story about a lecturing professor who claimed It isobvious
that ... and a student challenged him, saying, It is
notobvious. Then the professor paced back and forth for ten
minutes,thinking. Finally he announced, It IS obvious, and he
resumed hislecture.There is a corresponding story in music.
Nadia Boulanger assigned aharmonization exercise to a student,
from a book. She looked at hisresult for several minutes,
finally said, The alto part is not veryinteresting, and handed
it back to the student.David Ames===
===
Subject:
: Re: Graph Theory
TextbookIs the SpringerGTM Graph Theory by Reinhard Diestel
any good? Any suggestions arehighly welcomed.Yes definitely
great; BTW you can find an online copy of the book (googlewill
help you) so that you can make up your mind by yourself.Julien
Santini===
===
Subject:
: Re: Teaching philosophyteaching philosophy.
I think that gives me a good idea of the genre.What about the
length of the statement? Teaching Philosophy: Les fleurs qui
pousse de la merde by Allan Adler Evil is the root of all
money. Hence if funding has been foundfor a position, it
necessarily cometh of evil. Yet, though mushroomsnourish
themselves on excrement, they in turn nourish people and,
inlike manner, our fate as educators is to be a kind of
mushroom. Let metherefore describe some recipes in which this
mushroom has been used.[details omitted for the purpose of
this posting](I know, cut the first paragraph...)Allan
Adlerara@zurich.ai.mit.edu************************************
***************************************** ** Disclaimer: I am
a guest and *not* a member of the MIT Artificial **
Intelligence Lab. My actions and comments do not reflect ** in
any way on MIT. Moreover, I am nowhere near the Boston **
metropolitan area. **
**************************************************************
***************===
===
Subject:
: Re: Are the derivatives of
abs[(x-a)^3] different for x>a and x<a ?> given abs[(x-a)^3],
where abs[ ] means taking the absolute value, and> a is a
constant, are the first, second, third derivatives (with>
respect to x) different for x>a and x<a ?> I forgot how to
write down the derivative of an absolute function in a> formal
way, I mean, not using if else, but using something
likesign(x-a).The usual method is to define sgn(x) in terms of
the heaviside functionH(x) = { 0, x < 0; 1, x > 0; 0.5, x = 0
}, and write formallydH/dx = delta(x) where delta(x) = { 0, x
!= 0 } is the Dirac delta'function' (integral of delta(x) over
any interval containing 0 is 1 bydefinition, and 1/2 if 0 is an
endpoint).sgn(f(x)) = 2H(f(x)) - 1|f(x)| =
f*sgn(f)d(sgn(f(x))/dx = 2*delta(f(x))*f'(x)d(sgn(f)f)/dx =
sgn(f)df/dx + 2*f*f'*delta(f)Terms involving delta(f) vanish
if one avoids points where |f| = 0.> I think the first
derivative would look like:> 3(x-a)^2 * sign(x-a)Yes.> But
then I don't know how to proceed to the second derivative.You
have to exclude points where |f| = 0 from the domain to
proceedfurther; d/dx(delta(x)) is not well-defined.D(sgn(f)Df)
= sgn(f)(D^2)f + (Df)*(D(sgn(f))) (D = d/dx) = sgn(f) d^2f/dx^2
if |f| != 0.Similarly, D^n(sgn(f)f) = sgn(f)*(D^n)f if |f| !=
0.-- P.A.C. SmithThe vast majority of Iraqis want to live in a
peaceful, free world.And we will find these people and we will
bring them to justice.===
===
Subject:
: Re: JSH: Apology to Ramsay,
why I post> So I apologize to Keith Ramsay for questioning
his honesty here.> Why not apologise to those who did repost
that result for your perusal?Because that's obsessive
behaviour, and we all know that's wrong :-)V.-- email:
lastname at cs utk eduhomepage: cs utk edu tilde
lastname===
===
Subject:
: Re: oil of bitter almonds: Woehler and
Liebigpaper on Liebig's treatment of significant figures. I'll
take a lookat it.Allan
Adlerara@zurich.ai.mit.edu************************************
***************************************** ** Disclaimer: I am
a guest and *not* a member of the MIT Artificial **
Intelligence Lab. My actions and comments do not reflect ** in
any way on MIT. Moreover, I am nowhere near the Boston **
metropolitan area. **
**************************************************************
***************===
===
Subject:
: Re: the anticlassicalist }{ i:
linguistic negation <40325A2C.21936CE9@hate.spam.net>
<103513r3gafocff@corp.supernews.com>
<40328829.1E5C341D@hate.spam.net>
<1035esdt30h4d38@corp.supernews.com>
<m1g5UWHQ2zMAFwsg@baesystems.com>
<c11u1n$rcl$8@hercules.btinternet.com>
<NPQTaCDUYINAFwT1@baesystems.com>
<1039dr5kpjloif1@corp.supernews.com>In message
<1039dr5kpjloif1@corp.supernews.com>, galathaea >: >You should
have realised that anything full of -ism and -ist which
is>spewed>: >to sci.lang and sci.logic is not going to contain
any physics.>: >: (Nor language; I can't speak for sci.logic
;-)>:>: I'm well aware of that. What I was wondering was
whether the _OP_>: realises that there's a difference between
these isms and physics, and>: what it consists of.>Installment
vi >Before you lies the Void...>details the properties of the
closed linear subspace lattice>of a Hilbert space, detailing
the relationship with observables and>prediction in quantum
systems. This builds off of the mathematical theory>You do
understand the physical content, do you not?Haven't seen
any.Dirac compressed quantum mechanics into ~300 pages of
terse elegance, and didn't use ontology once.-- Richard
Herring===
===
Subject:
: Re: x^2 + y^4 = z^4> the equation x^2 +
y^4 = z^4 has no positive integer solutions.> Is the proof
... short enough for some kind soul to post it>An elementary
proof is based on descent. That is, assume a>solution exists
in positive x, y, z, and show that another one can be>found
with smaller z.> Are you sure?> That's true of x^4 + y^4 = z^2
where infinite descent is used at>
http://www.math.toronto.edu/mathnet/plain/questionCorner/
fermat4.html> However I see no way to modify the method used
there to this problem.descent methods for both x^4 + y^4 = z^2
andx^4 - y^4 = z^2 can be found in
:http://www.mathpages.com/home/kmath288.htm-- philippe(chephip
at free dot fr)===
===
Subject:
: Re: I got low score on math test,
please advise me and take a lookI would just stick to it. Any
way 77% is just about 85%. Just be sure tostudy things when
they come up and well now that you have had that test
youshould make it your job to get a full understanding of
every ting up to thatpoint. If you dont get something i am
sure there is some one around that canhelp you.Personally i
have only ever done subjects that are core to my degree.
orthat i wanted to do so i have never think about getting out
of it.In most uni's a score of 75-85 gives you a distinction.
which is good.maths is not about grades its about
understanding and fun! (i know its morethen that
aswell)stephen> my website states my case and has jpg files of
the four pages of the test> please take a look and advise me or
give me opinions> http://www.johncho.us===
===
Subject:
: Re: Pascals
TriangleThere was an interesting puzzle in the Sunday Times in
the UK recentlythatset me thinking about the issue that the
puzzle raised.The idea is that of a set of points arranged in
the familiar Pascal'striangle format (rows 1, 2, 3, ... etc.
containing 1, 2, 3, ... pointsin atriangular format) after
which the issue is that of the total number ofequilateral
triangles that can be found in this set of points.> Pascal's
triangle is> 1> 1 1> 1 2 1> 1 3 3 1> 1 4 6 4 1> ...> and does
not contain points, as such; it contains numbers.> Row 2 does
not contian a 2, but sums to 2. Row 3 does not contain a 3 or>
sum to 3.> You've completely lost me.I was using the Pascal
triangle analogy only to descibe the layout ofpoints:> .> . .>
. . .> . . . .That is, replace each of the numbers in the
Pascal triangle with a point andthen count the total number of
equilateral triangles (both 'up' and 'down'and all sizes) in
the resulting grid of points. The question then is 'whatis the
minimum number of points that need to be removed in order to
reducethe number fo such triangles to zero? Gladman===
===
Subject:
:
Re: Sets That Resemble Derivatives Somewhat[Snip: in
Doctorowese A' is the completement of A, A+B is the unionof A
and B and AB is the intersection of A and B]> 2) (A')' = A>
This doesn't correspond to differentiation,Yet another thing
that doesn't correspond to differentiationis that AB' + A'B
=/= (AB)'.Indeed I am at a loss reading these posts to find
any analogy betweencomplementation and differentiation (save
that Doctor Osherow usesthe same notation for both).--
===
===
Subject:
: Re: . The hardest of all hard facts .>Hi Peter,
You flip,> the Michelson-Morley experiment gave > results that
were non-null, > due to the effect of air in the apparatus
,>You're obviously wrong about that.Its true that Michelson
reported a null result.However recently the Process Physics
group at Flinders University inAustralia got hold of the data
from the Michelson-Morley experimentsand when they re-analysed
the data they concluded the result wasnon-null.They believe the
reason Michelson reported a null result was becausethe fringe
shifts he obtained were much smaller than he expected. Thiswas
due to the effect of the Lorentz contraction but because
Michelsonbelieved in Newtonian physics, he did not take this
into account.However, the Australian physicists found that
when they re-analysedthe Michelson-Morley data to take into
account the effects of both theLorentz contraction and the
reduced velocity of light through air,they obtained velocities
relative to space (or something in space)that correlated with
the rotation of the earth and its orbit round thesun.So they
claim that M-M type experiments can detect motion relative
tospace/ether/spacetime foam/quantum foam, whatever you wish
to call it,but there is an important proviso--The apparatus
must contain a gas,because if the experiment is done with a
vacuum interferometer, theeffect of the motion is perfectly
cancelled by the Lorentzcontraction.The re-analysis is
described in Michelson-Morley Experiments Revisited and the
Cosmic BackgroundRadiation Preferred
Framehttp://arxiv.org/pdf/physics/0205065The following paper
describes six other experiments that seem to havedetected
motion relative to space or (something in
space).http://arxiv.org/pdf/physics/0312082The above
re-analysis by the Australia physicists is partiallysupported
by an independent re-analysis by two Italian physicists.The
following abstract is from: The motion of the Solar System and
the Michelson-Morley experimentM. Consoli and E.
CostanzoIstituto Nazionale di Fisica Nucleare, Sezione di
Cataniahttp://arxiv.org/pdf/astro-ph/0311576---QUOTE---
Historically, the Michelson-Morley experiment has played a
crucialrole for abandoning the idea of a preferred reference
frame, theether, and for replacing Lorentzian Relativity with
Einsteins SpecialRelativity. However, our re-analysis of the
Michelson-Morley originaldata, consistently with the point of
view already expressed by otherauthors, shows that the
experimental observations have beenmisinterpreted. Namely, the
fringe shifts point to a non-zeroobservable Earths velocity
Vobs = 8 4 0 5 km s. Assuming theexistence of a preferred
reference frame, and using Lorentztransformations to extract
the kinematical Earths velocitythat corresponds to this Vobs ,
we obtain a real velocity, in theplane of the interferometer,v
earth = 201 12 km s. This value is inexcellent agreement with
Millers calculated value Vearth = 203 8km/s and suggests that
the magnitude of the fringe shifts isdetermined by the typical
velocity of the Solar System within ourgalaxy. This conclusion,
which is also consistent with the results ofall other classical
experiments, leads to an alternative interpre-tation of the
Michelson-Morley type of experiments. Contrary to thegenerally
accepted ideas of last century, they provide
experimentalevidence for the existence of a preferred
reference frame. This pointof view is also consistent with the
most recent data for theanisotropy of the two-way speed of
light in the vacuum.--ND QUOTE---The following paper proposes
that motion relative to space (orsomething in space) should be
detectable, because space is full ofVacuum condensates and
ether-drift experimentsM. Consoli, A. Pagano and L.
PappalardoIstituto Nazionale di Fisica Nucleare, Sezione di
Cataniahttp://arxiv.org/pdf/physics/0306094The following is
also interesting.Modern Michelson-Morley experiments and
gravitationally-inducedanisotropy of cM. ConsoliIstituto
Nazionale di Fisica Nucleare, Sezione di
Cataniahttp://arxiv.org/pdf/gr-qc/0306105Peter===
===
Subject:
: Re:
I got low score on math test, please advise me and take a
look>my website states my case and has jpg files of the four
pages of the test>please take a look and advise me or give me
opinionsWell, before looking at the test I see whining about
how you shouldbe allowed to drop after the drop date because
it took more than48 hours for the instructor to return the
test. My opinion on thisis (i) asking us for our opinion on
policies at your school is utterlystupid, (ii) whining about
the policies is stupid. _Is_ there a rulesomewhere stating
that tests administered shortly before thedrop date must be
returned by the drop date? I didn't think so._Did_ the
instructor promise to return the test by the drop date?I
didn't think so.I don't know about the rules where you are,
but _here_ there_is_ a rule that states explicitly the test
wasn't returned on timeshall _not_ be a valid reason to be
allowed to drop past theofficial date.Good luck with that.Now
glancing at the test:Can't see page 1: I get a page-size jpg,
but it's blank exceptfor the to inch or so. Same problem with
page 2.I can see page 3. First thing I notice is the complaint
aboutneatness - expecting someone to be able to read that
sortof scrawl is totally unreasonable. Let's see:problem 8:
9.75 points out of 10. I'm not sure why he tookoff a quarter
point, possibly because he couldn't read it.Let's say to be
generous you have another 1/4 point coming there.problem 9:
4.5 out of 5. I would have given you fewer pointsthan that,
because there's a bit of explanation missing(you need to point
out that the function is not even_defined_ at the point in
question, because thatdenominator vanishes). Also your sketch
is _not_ whatthe function looks like - where did that extra
humpcome from? I'd give you at _most_ 3 points, probably
less;that 1.5 down, so my score is 1.25 below what you gotso
far.problem 10: 1 out of 5. You have 0 out of 5 coming onthat
one - it _said_ using properties of limits and Idon't see a
single property of limits mentioned, anywherein your solution.
So far my estimate is 2.25 below what you got.problems 11, 12:
you got full credit.problem 13: you missed 1.25 points. I
can't quite tell fromthe jpg, it looks like he took off 1/4
point for writing f'(x)where you meant f'(a). I can't see
where the other pointdeducted is, but I would have taken off
more than aquarter point for the error I see, so we'll call
this one even.problem 14: It looks like you lost 1.5 points
for sayingv'(t) = -2. What you said is wrong; v'(t) = 2t - 6,
it'sv'(2) that equals -2. 1.5 points seems reasonable to
me.Well there you have it. I can only see half the test; onthe
half that I can see you probably would have got afew points
_less_ than you did if I'd been grading the test.My advice is
next time be more careful about the details.And when you're
asking to use your calculator to sketcha graph and you see a
parabola draw something that_looks_ like a parabola! I mean on
that one you justhad to copy what you saw on the
screen...>http://www.johncho.us************************===

===
Subject:
: Re: JSH: Splitting field, algebraic integer factors>
Previously I posted that if you can't *see* the factors
between> irrational algebraic integers then they're not there,
but more> correctly the situation is that two algebraic
integers have to be> members of the same splitting field to
have a non-unit algebraic> integer in common.> (I say more
correctly as there may be some terminology issues here>
because what mathematicians currently call a splitting field
is not> a true field, but something close, like the field of
rationals. But> that's another issue for another time.)>
That's a nifty and powerful result. Why am I the one who had
to> discover it?Well I was wrong. I should have known in
retrospect, but it was yetanother situation where I kind of
decided that I wanted something andthought I had it, only to
find out later I was wrong.There's no point to hanging on to
wrong things though.Luckily, mathematics is supposed to be an
arena where the truthmatters.===
===
Subject:
: Re: Sets That
Resemble Derivatives Somewhat>[Snip: in Doctorowese A' is the
completement of A, A+B is the union>of A and B and AB is the
intersection of A and B]> 2) (A')' = A> This doesn't
correspond to differentiation,>Yet another thing that doesn't
correspond to differentiation>is that >AB' + A'B =/=
(AB)'.>Indeed I am at a loss reading these posts to find any
analogy between>complementation and differentiation (save that
Doctor Osherow uses>the same notation for both).Evidently
you've forgotten that derivatives satisfy (f + g)' =
(f')(g').************************===
===
Subject:
: Re: strain
softening spring>... I posted a file> called DE.pdf showing
the solution at:> http://math.asu.edu/~kurtz/de/> --LynnThank
you Lynn! This is most helpful - I would never have
figuredthis out by myself in a zillion years!===
===
Subject:
: Re:
strain softening spring> Thank you very much Michael. I will
try out these approaches. Theintuitive approach appeals to
me.sincerelyPaul===
===
Subject:
: Re: Quadratics and transformation>
It seems to me that I should be able to reasonably easily get
a> general idea of the type(s) of transformations that have
been applied> to x^2 just by looking at the finished
quadratic. > That is, if I have 3x^2 + 7x + 2, can I easily
relate the coefficients> and constant to the dilations,
reflections and translations that this> function represents
compared to x^2 ? > I have tried thinking about it as a
combined function, that is,> f(x)=x^2, g[f(x)]=3x^2 + 7x + 2
and trying to find g(x). Well, I> haven't tried with this
example but one I tried earlier turned into a> horrible mess
and didn't give me what I wanted - I assumed that g(x)> is
simply the inverse of g[f(x)] so maybe that's why it got so
messy ?> It also seemed to be a lot of work!Transform both x
and y First by scale then the translations> I'm aware that I
could determine the turning point and work out the>
translations and that the roots would indicate what
horizontal> dilation has been applied. I expect that if I
calculate two other> points I would also be able to determine
reflection and vertical> dilation. Of course I could always
graph it but I'm hoping to achieve> what I want from
inspection of the description of the function (is> that the
right terminology?). > Any and all help appreciated.> Ivan.If
you only look at parabolas with v=u^2 as your ideal(1)
Ax^2+Bx+C=y formBy inspectionA Scale A>0 positive orientation,
i.e. parabola faces upA<0 negative orientation, i.e. parabola
faces DownB/2A Horizontal translation C/A-B^2/4A^2= Vertical
Translation{B^2-4AC} :: DiscriminantQuadratic formulax =
[-B+/-(B^2-4AC)^(1/2)]/2ALet u=U(x) and v=V(y)u^2=vFind U(x)
And V(y)(1) Ax^2+Bx+C=y Divide by A A<>0(2) x^2 +(B/A)x+(C/A)=
y/ASubtract C/A from both sides(3)x^2 +(B/A)x = (y-C)/AComplete
the Square (add (B/2A)^2 to both sides(4) x^2 +(B/A)x+B^2/4A^2=
(y-C)/A+B^2/4A^2(4a)(x+B/2A)^2= (y-C)/A+B^2/4A^2LHS = {U(x)^2}
RHS= V(y)(5) u=(x+B/2A) v=(y/A+(B^2-4AC)/4A^2)Standard
form(x-h)^2=4p(y-k)Vertex (h,k) focus (h,k+p)Axis x = h
directrix y= k-pGeneral formAx^2+Bx+C=y soh = -B/2A p = 1/4Ak
= (B^2-4AC)/4ABe careful when substituting A,B & C are all
signedCarl-- If its Monday then I am a fool but not
ignorant.===
===
Subject:
: Re: . The hardest of all hard facts
.>They found none. Miller's results could not be duplicated by
anyone but>Miller with his apparatus.The following paper
describes a number of experiments that seem tohave detected
motion relative to space or (something in
space).http://arxiv.org/pdf/physics/0312082The following
abstract is from: The motion of the Solar System and the
Michelson-Morley experimentM. Consoli and E. CostanzoIstituto
Nazionale di Fisica Nucleare, Sezione di
Cataniahttp://arxiv.org/pdf/astro-ph/0311576---QUOTE---
Historically, the Michelson-Morley experiment has played a
crucialrole for abandoning the idea of a preferred reference
frame, theether, and for replacing Lorentzian Relativity with
Einsteins SpecialRelativity. However, our re-analysis of the
Michelson-Morley originaldata, consistently with the point of
view already expressed by otherauthors, shows that the
experimental observations have beenmisinterpreted. Namely, the
fringe shifts point to a non-zeroobservable Earths velocity
Vobs = 8 4 0 5 km s. Assuming theexistence of a preferred
reference frame, and using Lorentztransformations to extract
the kinematical Earths velocitythat corresponds to this Vobs ,
we obtain a real velocity, in theplane of the interferometer,v
earth = 201 12 km s. This value is inexcellent agreement with
Millers calculated value Vearth = 203 8km/s and suggests that
the magnitude of the fringe shifts isdetermined by the typical
velocity of the Solar System within ourgalaxy. This conclusion,
which is also consistent with the results ofall other classical
experiments, leads to an alternative interpre-tation of the
Michelson-Morley type of experiments. Contrary to thegenerally
accepted ideas of last century, they provide
experimentalevidence for the existence of a preferred
reference frame. This pointof view is also consistent with the
most recent data for theanisotropy of the two-way speed of
light in the vacuum.--ND QUOTE---The following paper proposes
that motion relative to space (orsomething in space) should be
detectable, because space is full ofVacuum condensates and
ether-drift experimentsM. Consoli, A. Pagano and L.
PappalardoIstituto Nazionale di Fisica Nucleare, Sezione di
Cataniahttp://arxiv.org/pdf/physics/0306094The following is
also interesting.Modern Michelson-Morley experiments and
gravitationally-inducedanisotropy of cM. ConsoliIstituto
Nazionale di Fisica Nucleare, Sezione di
Cataniahttp://arxiv.org/pdf/gr-qc/0306105The abstract
states:The recent, precise Michelson-Morley experiment
performed by Muller etal. suggests a tiny anisotropy of the
speed of light. I propose aquantitative explanation of the
observed effect based on theinterpretation of gravity as a
density fluctuation of the Higgscondensate.Peter===
===
Subject:
:
Call for participation (Omphalos)(Apologies if you receive
this message more than
once)---------------------------------------------------Intro-
----Omphalos is the first context-free language learning
competition thattask is to infer a model of a context-free
language from unstructuredexamples (both positive and
negative) and to use that model to label aset of test
sentences as being either in or out of the language.The
Competition Task--------------------We have generated some
context-free grammars. For each grammar we havelabeled a set
of sentences indicating whether or not those sentencescan be
generated from that grammar. The goal of the competition is
toinfer a model of each language (such as a grammar) using the
trainingdata. You then need to tag new sentences, indicating
whether thesentences are in the language or not.
---------------More Information----------------For more
information, check the website:http://www.irisa.fr/Omphalos/or
contact the organizers at:omphalos@irisa.frOmphalos is being
organized and administered by:Brad Starkie, Fran?ois Coste,
and Menno van Zaanen ===
===
Subject:
: Re: errors in an argument>
My favorite is birds : How can one evolve into flying ? So
many> things have to happen in order to create functional
wings it seems> impossible any awkward animal developing in
this direction would> survive it's cost, or any failing
flying attempts. The evolutionist> arguments sound totally
absurd : they ran fast or jumped from tree> to tree and got
an advantage slowly developing wings. Sounds like> absolute
voodoo to me. >That depends on the size of the animal. If the
animal is small>enough, the terminal velocity can result in a
non-lethal impact with>the ground. (I believe the threshold is
around the size of a mouse or>a cat.) Incremental levels of
control over the landing point then>have an obvious benefit.
>-- >Daniel W.
Johnson>panoptes@iquest.net><a>href=http://members.iquest.net/
~panoptes/>http://members.iquest.net/~>panoptes/</a> 039 53 36
N / 086 11 55 W> That's highly unlikely. As far as I know the
first flying creatures> they talk about are dinosaurs, no
less.20to 24 size the big ones came along a lot later And
quite big ones as well. In> any case, a functional wing on a
mouse, that allows even such a> manuver is still something
very hard to produce simply by mutation.Two mutation produce
flaps of skin between digits, goes back to the Fish days other
Transforms also in the genome from the fish ansestor > This is
a very hard one to crack for the theory of evolution. I still>
didn't hear a good solution for it.-- If its Monday then I am a
fool but not ignorant.===
===
Subject:
: For what purpose ?Hi Peter,
You mentioned, recently the Process Physics group at Flinders
University in Australia got hold of the data from the
Michelson-Morley experiments and when they re-analysed the
data they concluded that the result was non-null .Why
re-analyse the data from such an ancient and flawed experiment
?The Michelson-Morley experiment was good enough to suggest
that the speed of light in a vacuum is the same for all
observers. ( Which it is, to a very high precision )The
Kennedy-Thorndike experiment ruled out the Fitzgerald-Lorentz
contraction hypothesis that objects experienced a compression
in the direction of their motion relative to some third frame
of reference.You suggest introducing a third frame of
reference into relativity ... For what purpose ?===
===
Subject:
:
Solving Equation with lnI was working on a math problem at
work and ran into something I don'tknow how to solve and was
hoping someone could give me a clue on thisfairly simple
problem.Original Equation:227.11 = 5.8{ln(38/x)} +
(38-x)Simplfied Equation:189.11 = 5.8*{ln(38/x)} - xof course
you can reduce this down to32.60 = ln(38/x) - x/5.8and further
(I think) to 1.44 x 10^14 = 3.8/x - e^(x/5.8)but I am stuck
after this. Is it possible to solve for X?===
===
Subject:
: Re: I
got low score on math test, please advise me and take a look>
my website states my case and has jpg files of the four pages
of the test> please take a look and advise me or give me
opinions> since I took the test last Monday I feel 48 hours
was ample time to > return the test beck on Wednesday
especially with the withdraw without > a w date near.
Furthermore the test could have been made available by >
Tuesday, that is one week and one day from the test date.> I
feel I should be allowed to withdraw without a W under these >
circumstances.Frankly, you're wrong on a number of levels.It
would be a matter of courtesy if the professor would get
theexam back quickly so that these decisions could be made,
buthe is under no professional obligation to do so, and 48
hoursis not a very long time. Here's a few issues:1. It's not
fair if wealthier students purchase higher GPA'sbecause they
can afford to take lower course loads and spend6 years getting
a 4-year degree. It's even more unfair ifthey get to spend
several weeks test driving their coursesand throw away the
ones they are getting bad grades in.It's even more worse that,
as you are doing, they throw awaythe courses they are getting
_average_ grades in. This isthe main reason there is a W
deadline. Probably you hadother feedback upon which to make
your decision.2. Education is not about working the system to
manufacturethe highest possible GPA, but about learning and
maturing. TheGPA is a (rough) measurement of how much learning
and maturinghas taken place. If you succeed in beating the
system byartificially elevating your GPA, you won't have any
idea whether you've learned anything or matured properly. 3.
It takes a long time to evaluate an exam properly. I
knowprofessors who grade (and brag about it) stacks of 70 or
100math exams overnight, but watching them in action, the
techniqueis strictly slash and burn. I usually take a week,
even ifthe stack is only 10 or 20 high. I read the exams for
patternsin the overall understanding of the class to see if I
need togo back and re-cover something, and I think about other
adjustmentsto the curriculum. Besides evaluating each student,
I need toevaluate myself and the flow of the course.I would
like you to notice two things about your own case:A: You
complain that the test was not evaluated accurately.B: You
complain that it was not evaluated quickly enough.These are
very contradictory wishes. 4. In the course of my career, I've
had the un-pleasure of havingto proctor math placement exams to
incoming freshmen. These testsare multiple-choice, fill in the
bubble and the only result of thetests is to place students in
Basic Math / Intermediate Algebra /College Algebra /
Precalculus / Calculus. The idea is to evaluatethe preparation
of the student and put him in the course he's ready for. For
some crazy, unknown reason, these 18-year-olds cheat on
thesetests. There's no GPA involved here. The only penalty for
successfully cheating is to be placed in a math course which is
too hard for you. It seems to me that a person of such low
character as would cheat, would also figure out that the
thingone wants to do in this situation is to purposely get a
low score in order to be placed in an easy course and get a
cheapA. But every Summer, there they sit, with their baseball
capspulled down over their eyes trying to peek across the
aisle........and I just let 'em.And the point of this nearly
irrelevant story is that sometimesa guy needs to sit back and
blink a couple of times and saywhat am I doing here? You're
taking Calculus for a reasonand you got a test score that was
lower than you wanted. I'ma bit surprised that your immediate
_reflex_ is to drop the class.Most students' immediate reflex
would have been to begin studyingharder. What is the reason
you are taking Calculus? If that reason isimportant (such as
part of an engineering major) then puttingit off is likely a
bad idea. My story is supposed to show thathigh school
students have a sort of mentality about things whichcontrols
their decisions. It's important to get high scores _period_.
So I'll cheat on the placement test. Likewise,undergraduates
have some kneejerk reactions to situationsthat cause them to
make decisions without thinking.You think you deserved an 84
on the test, but you got a 77. Ihave very little data to work
with, but I can't help but thinkthat you could make up the 7
points and not drop the class atall.===
===
Subject:
: Re: errors in
an argument>I recently attended some talks by a creationist
talking about>how bad evolution is.<snip>Anyway, at one point
in his presentation, he attempted>to prove that life could not
arise by chance. Yes, creationist do that a lot. They tend to
incorperate a lot ofthings which have little to do with
evolution in their 'evolution isbad, so we must be right'
speeches. In this case he was trying to show that abiogenesis
was impossible.The method he chose is not new. In fact it is
so standard that thereis
this:http://www.talkorigins.org/faqs/abioprob/===
===
Subject:
: Re:
A newbie's question -- about real numberX-Cise:
tanbanso@iinet.net.auX-CompuServe-Customer: YesX-Coriate:
admin@interspeed.co.nzXcrate: tanandtanlawyers.comX-Pose:
george_cox@btinternet.comX-Punge: Micro$oftX-Sanguinate:
themvsguy@email.comX-Terminate: SPA(GIS)X-Tinguish: Mark
Griffith <markgriffith@rocketmail.com>X-Treme: C&C,DWS at
07:08 PM, chergarj@cs.comhaho (Chergarj) said:>Theoretically
NO.What theory are you using? His r2 is 0.88 + sigma^i=3_00
9*10^{-i},and sigma^i=3_00 9*10^{-i} converges to 10^{-2} =
0.01, so r2 = 0.88 +0.01 = 0.89, which is his r1.-- Shmuel
(Seymour J.) Metz, SysProg and JOATUnsolicited bulk E-mail
will be subject to legal action. I reservethe right to
publicly post or ridicule any abusive E-mail.Reply to domain
Patriot dot net user shmuel+news to contact me. Donot reply to
spamtrap@library.lspace.org===
===
Subject:
: Re: tensors for
totsX-Cise: tanbanso@iinet.net.auX-CompuServe-Customer:
YesX-Coriate: admin@interspeed.co.nzXcrate:
tanandtanlawyers.comX-Pose: george_cox@btinternet.comX-Punge:
Micro$oftX-Sanguinate: themvsguy@email.comX-Terminate:
SPA(GIS)X-Tinguish: Mark Griffith
<markgriffith@rocketmail.com>X-Treme: C&C,DWS>How far can one
get in understanding the structure of tensors at a>point on a
manifold in terms of our old school friends column>vectors,
row vectors and matrices?You'll get bogged down fairly
quickly.>Even in general relativity all we have at each point
on the >manifold is a real vector space with four
components,No. You also have, e.g., g and R. You could write
the components of gas a matrix, but that would be horribly
misleading. R is a lost cause;you really need to forget about
matrices and just think of it as atensor. at 02:28 PM,
nulldev00@aol.com (Edward Green) said:>Or for that matter, say
we want v* to>be the dual vector corresponding to vYou need a
metric in order to do that.>orthogonal
transformationsMeaningless without a metric.>I also hope not
to get hung up on any terminological problems>between math and
physics, perhaps by saying things like we'd like>this object to
be invariant, instead of using terms like>covariant and
contravariant, to start.First, you'd need to define what you
mean by invariant. Second, youwon't get very far if you don't
distinguish covariant fromcontravariant, and both from
mixed.>Anyway, to take one more baby step, if I have not
exhausted my>credits, am I correct in thinking that a
reasonable way to introduce>a metric tensor into babyland
would be to define the inner product>of a row vector and a
column vector, in a particular coordinate>system, to be a
bilinear form with a matrix sandwiched in the>middle, that
matrix our metric?No, you don't need a metric for that. You
need a metric tensor todefine the inner product of two row
vectors, the inner product of twocolumn vector and the
transpose operation that takes row vectors intocolumn vectors
and vice versa.-- Shmuel (Seymour J.) Metz, SysProg and
JOATUnsolicited bulk E-mail will be subject to legal action. I
reservethe right to publicly post or ridicule any abusive
E-mail.Reply to domain Patriot dot net user shmuel+news to
contact me. Donot reply to
spamtrap@library.lspace.org===
===
Subject:
: Re: Solving Equation
with ln> I was working on a math problem at work and ran into
something I don't> know how to solve and was hoping someone
could give me a clue on this> fairly simple problem.> Original
Equation:> 227.11 = 5.8{ln(38/x)} + (38-x)> Simplfied
Equation:> 189.11 = 5.8*{ln(38/x)} - x> of course you can
reduce this down to> 32.60 = ln(38/x) - x/5.8> and further (I
think) to > 1.44 x 10^14 = 3.8/x - e^(x/5.8)> but I am stuck
after this. Is it possible to solve for X?> Not in terms of
the standard elementary functions. But youcan solve in terms
of the Lambert W function...x =
W(38*exp(-(189.11)/5.8)/5.8)*5.8 =
2.627466197*10^(-13);===
===
Subject:
: unsolved constrained
correlation problemI posted originally the following problem
on sci.stat.math,but I received solutions working only for the
independant case.We consider two random vectors X and Y, having
the same distribution,such that the variance matrix V is
diagonal. The common distributionis fixed, not the random
vectors.For simplicity, we assume that X and Y are vectors in
the plane.It is pointed out that r(X1,X2)=0 but X1 and X2
could be dependant.We consider a joint distribution of (X,Y)
such that r(X1,Y1)=0 andr(X2,Y2)=1. Does such a joint
distribution always exists and why ?Michel Petitjean, Email:
petitjean@itodys.jussieu.frITODYS (CNRS, UMR 7086)
ptitjean@ccr.jussieu.frhttp://petitjeanmichel.free.fr/
itoweb.petitjean.html===
===
Subject:
: Re: Collatz Conjecture :
Symmetry question.> I have a program that can build each level
from the previous one.I also have written this program. I am
using an arbitrary precisionlib to allow large numbers. I have
two applications, one calculatesjust end points, and the other
traverses the tree. We should comparenotes and maybe work
together and release a simple tool for others touse. Just a
thought.===
===
Subject:
: Nonlinear PDE HelpHere's a partial
differential equation that I'm almost positivehas a unique
solution, although I don't have any idea how to goabout
finding it. Does anyone have suggestions?We want to find
f(x,y) subject to the following two conditions 1. f(x,0) =
g(x) 2. (df/dx)^2 + (df/dy)^2 = 1where g(x) is some known
differentiable function of xand where d/dx and d/dy are
supposed to be partial derivatives.I know of a solution for a
specific case: g(x) = square-root(1 + x^2).Then a solution is
f(x,y) = square-root(x^2 + (y+1)^2)However, I don't know how
to leverage this particular solution toget a solution for an
arbitrary g(x).===
===
Subject:
: Re: errors in an argument> I admit
that my knowledge of the research of small dinosaurs isn't
very> deep, so you may be right, but that keeps my general
point intact.A little more research turns up a wingspan of 0.5
m for Archaeopteryx.That puts it at the approximate size of a
housecat, a critter known tosometimes survive a fall even if
it reaches terminal velocity. (Notethat sometimes survive is
an ideal condition to drive naturalselection.)> The way it
seems to me (and I'm not talking as an expert in aerodynamics>
either) is that even for extremely small and light creatures,
the ability> to develop a functional limb that will allow it
even the smallest> maneuverability by mutation benefit is
unlikely. Remember that it has to> be aerodynamic to some
extent, and to be evolutionary beneficial. It's> very hard for
me to imagine how such a thing would look like.Imagine
this:http://www.animalnetwork.com/critters/profiles/
flyingsquirrel/default.aspKeep in mind that embryos of many
species typically have webbing duringdevelopment; it would not
take much of a mutation for the webbing topersist.A little
research also turns up an explanation of feathers that
hasnothing to do with their eventual use in flight. --
http://members.iquest.net/~panoptes/039 53 36 N / 086 11 55
W===
===
Subject:
: equivalent expressionsI'm sure this is just a
print error, but a Foundations book I haveasks the
following:[Q] Give an example of two equivalent
[propositional] expressions thatdo not have the same truth
table.Is there something deep I'm missing here, because what I
see is acontradiction. Perhaps my misunderstanding lies in what
they mean bytruth table.Any help is greatly
appreciated.R===
===
Subject:
: Re: Collatz Conjecture : Symmetry
question.http://arxiv.org/abs/math.NT/0312309I think it's
clear that it will never be proven, because there is justnot
enough time to prove it.Craig===
===
Subject:
: Re: Got a speeding
ticket and need to fight backwanted-suresh> Good one, chief. I
hear the WB network is hiring comedy writers.WB is also looking
for lawyers who can play sidewalk Santas for theirchristmas in
july specials. Btw, WB is also looking for statistcianswho can
prove that christmas is supposed to start in july.> It took you
six days to come up with *that*? Pathetic.> Doug===
===
Subject:
: Re:
To users of GMP> The FFT code in GMP currently contains a flaw
that leads to> inaccurate results.Can you please give some
order of magnitude for the number of bitsbelow which it can be
safely assumed that the bug will not occur(maybe just because
the FFT code is not even called).> More information can be
found at http://www.swox.com.Could not find more thanA bug in
the FFT multiply code that can cause miscomputation has been
found. Until we can provide a fix, and until we have performed
extensive further testing of the code, all users are urged to
recompile GMP using the configure option --disable-fft.TIA,
Francois Grieu===
===
Subject:
: smallest eigenvalue of
Laplacianthere is a theorem that says that an eigenfunction
corresponding tothe smallest eigenvalue of the Laplacian on
some bounded domain has nozeros inside that domain.Is the
reverse implication also true? Meaning: Does
everyeigenfunction without zeros inside the domain belong to
the smallesteigenvalue?Thank you in advance for any helpful
comments.Yours sincerely,Tobias N.8ahring-- See me at
tn-home.de/Tobias/===
===
Subject:
: Re: If We Replaced Each Prime
With -1...> Let c(k) = the sum of the exponents in the prime
factorization of k.> So, if we exchanged each prime in the
prime factorization of k with> (-1),> we would get (-1)^c(k).>
What I am wondering, however, is> what is C(m) => sum{k=1 to m}
(-1)^c(k)> asymptotical towards?The function (-1)^c(k) is
usually denoted lambda(k) and is known asLiouville's lambda
function. Its partial sums (up to m) are o(m); thisis
equivalent to the prime number theorem. The stronger statement
thatits partial sums are O(m^{1/2+epsilon}) for each fixed
epsilon > 0(and m->oo) is equivalent to the Riemann
hypothesis.You may have seen the same results stated for the
Mobius function. Onecan relate the partial sums of lambda to
those of mu and thereby goback and forth between these results
as follows. As you note,> C(m) also equals:> sum{k=1 to m}
floor(sqrt(m/k)) *mu(k),and this can be put in the alternate
form C(m) = sum_{k <= sqrt(m)} M(m/k^2),where M(x):=sum(mu(n),
n<=x); one can also verify the related identity M(m) = sum_{k
<= sqrt(m)} mu(k) C(m/k^2). and similarly for the estimate
mentioned in connection with theRiemann hypothesis. On the
other hand, neither C(m) = o(m^{1/2}) orM(m) = o(m^{1/2}) can
hold (as m->oo); this is because either wouldimply zeta had no
zeros on Re(s) = 1/2.You may be interested in Chapter 4 of my
online notes which presentsan elementary proof of the PNT (in
the form M(x) = o(x) as x->oo) andwhich discusses some of
these points. See
http://www.princeton.edu/~ppollack/notes/Hope this
helps,Paul===
===
Subject:
: Re: Teaching philosophy> I need to write
a description of my teaching philosophy. In order to> do so
accurately, I think I need to write at greater length and in>
greater detail on this topic than any hiring committee will
want to read.> Moreover, I need to present my views accurately
but, somehow, in such> a manner as not to vitiate the
consideration of my application.teaching philosophy a
significant number of mathematicians wouldembrace:teach with
depth and require students to learn by setting
highstandards.teaching philosophy majority of bureaucrats and
managers plagueingacademic institutions want to see:full
commitment to teach in a diverse environment with
provenexcellence in teaching. (whatever the hell that
means...)teaching philosophy majority of mathematicians
practice:teach one or two sections per class in order to meet
an extensivenon-realistic curriculum realised by either a
bureaucratic committteeor a dictator.> That being the case,
can someone please tell me what my teaching> philosophy is?>
Ignorantly,> Allan Adler> ara@zurich.ai.mit.edu>
**************************************************************
**************> * *> * Disclaimer: I am a guest and *not* a
member of the MIT Artificial *> * Intelligence Lab. My actions
and comments do not reflect *> * in any way on MIT. Moreover, I
am nowhere near the Boston *> * metropolitan area. *> * *>
**************************************************************
**************===
===
Subject:
: Matrix math -- verify
please?explicitly stated, so I wanted to verify:- The product
of two rectangular matrices (e.g. A = 4x3 and B = 3x4)
cannotbe invertible, since each matrix has at most rank 3.-
The product AB has at most rank 3. (But can be less)- Even if
A has rank 3 and B has rank 3, AB may still have rank <
3.===
===
Subject:
: Re: To users of GMPCc: Francois Grieu
<fgrieu@micronet.fr> The FFT code in GMP currently contains a
flaw that leads to> inaccurate results.> Can you please give
some order of magnitude for the number of bits> below which it
can be safely assumed that the bug will not occur> (maybe just
because the FFT code is not even called).The file gmp-mparam.h
contains thresholds concerning which algorithms willbe used.
Look for MUL_FFT_THRESHOLD, which will be in the thousands,
andmultiply by the number of bits per limb, which will be 32
or 64 depending onyour machine word size. That will be the
number of bits below which the FFTcode will not be called.>
More information can be found at http://www.swox.com.> Could
not find more thanA bug in the FFT multiply code that can
cause miscomputation> has been found. Until we can provide a
fix, and until we have> performed extensive further testing of
the code, all users are> urged to recompile GMP using the
configure option --disable-fft.> TIA,> Francois GrieuThere is
a new random number generator developed which can be found
atandomb.objwhich makes the span lengths proportional to the
size of the output.More information about the random code can
be found at.The new random code triggered the bug in the FFT
code almost immediatelyafter testing began.===
===
Subject:
: Induced
completeness to stronger metric spacesTake a nonmpty set X and
two metrics d1(x,y) and d2(x,y) in X. Calld2 stronger than d1
if:d1(x,y) <= C d2(x,y) forall x, y in X, where 1 <= C < ooLet
d2 be stronger than d1.It's easy to see that if (x_n) is a
Cauchy sequence in (X, d2) thenit's also Cauchy in (X, d1).
Now, assume (X, d1) is complete in thesense that every Cauchy
sequence converges to some x in X. Does thisimply that (X, d2)
is also complete?Counterexamples would be welcome.-- I'm not
interested in mathematics that might have anythingto do with
reality. -- Russell Easterly, in sci.math===
===
Subject:
: Re: x^2 +
y^4 = z^4===>
===
Subject:
: Re: x^2 + y^4 = z^4> the equation x^2
+ y^4 = z^4 has no positive integer solutions.> Is the proof
... short enough for some kind soul to post it>An elementary
proof is based on descent. That is, assume a>solution exists
in positive x, y, z, and show that another one can be>found
with smaller z.> Are you sure?> That's true of x^4 + y^4 = z^2
where infinite descent is used at>
http://www.math.toronto.edu/mathnet/plain/questionCorner/
fermat4.html> However I see no way to modify the method used
there to this problem.> ----The impossibility of x^4 + y^4 =
z^4 follows immediately from theimpossibility of x^4 + y^4 =
z^2 since z^4 = (z^2)^2, i.e. a solutionto the first equation
necessarily gives one to the second. Thedescent is used for
x^4 + y^4 = z^2.===
===
Subject:
: Re: Graph Theory Textbook> I'm an
undergraduate, and this summer I will be participating in an>
REU for discrete math and combinatorics. I am looking for a
good> graph theory textbook to learn the basics from. I have
seen the books> that Dover (which of course are very cheap)
has to offer on the> subject, but unfortunately they seem too
elementary. Is the Springer> GTM Graph Theory by Reinhard
Diestel any good? Any suggestions are> highly welcomed.> JackI
like Bondy and Murty. Also Harary's book is a classic. You
mightalso consider Graver and Watkins _Combinatorics with and
Emphasis onthe Theory of Graphs_. For applications Fred
Roberts's _DiscreteMathematical Models_ is fun.===
===
Subject:
: Re:
Are the derivatives of abs[(x-a)^3] different for x>a and x<a
?>You have to exclude points where |f| = 0 from the domain to
proceed>further; d/dx(delta(x)) is not well-defined.Um, you
just derivated the distribution related to the Heaviside
stepfunction. How can derivating the Dirac delta distribution
suddenly notbe defined?-- I'm not interested in mathematics
that might have anythingto do with reality. -- Russell
Easterly, in sci.math===
===
Subject:
: Re: I got low score on math
test, please advise me and take a look by
support1.mathforum.org (8.11.6/8.11.6/The Math Forum,
$Revision: 1.9 primary) id i1JHEsb01554;> my website states my
case and has jpg files of the four pages> of the test please
take a look and advise me or give me> opinionsthere is little
doubt that your teacher is an imbecile.prob 1 deserves more
credit.prob 3 deserves full credit.prob 6a deserves more
credit.prob 6b deserves no credit.prob 8 deserves full
credit.prob 10 deserves no credit.prob 13 deserves more
credit.btw, you may want to tell your techer to give you more
timeso that you can write neater:)><a
href=http://www.johncho.us>http://www.johncho.us</a>===
===
Subject:

: Re: Axioms defining a finite field===>
===
Subject:
: Axioms
defining a finite field>Let (F, +, *) be a finite set with two
operations>and constants 0, 1 such that the following rules
hold:>(1) a + (b + c) = (a + b) + c>(2) a + 0 = a>(3) for
every a there's a b so that a + b = 0>(4) a*(b*c) =
(a*b)*c>(5) a*1 = a>(6) 1 is distinct from 0>(7) a*(b + c) =
a*b + a*c>(8) (a + b)*c = a*c + b*c>(9) a*b = 0 => a=0 or
b=0>Show that F is a field. Can one of the rules be omitted>so
that F still has to be a field?>let a /= 0, b0 = 0, b1 = 1>A =
{ a bj | 0 <= j <= |F| }>|A| = |F| because> if a bj = a bk:
a(bj - bk) = 0; bj - bk = 0; bj = bk>As F is finite, 1 in A>
which shows a has right multiplicative inverse a_r.>This with
a1 = a, shows F0 is a group under *.>Did you forget a+b = b+a,
ab = ba?(1) - (3) imply F is a group under +Next show a*0 = 0
for all a. This follows as usual from a*0 = a*(0 + 0).Then
from (9) and (7), you can deduce that, for a in F - {0}, a*b =
a*cimplies b = c (just add a*(-c) to both sides), and hence, by
finiteness,there exists a^-1 with a*a^-1 = 1. Then, from (4)
and (5), F - {0} is a group under *.Now we can use (7) and (8)
to deduce a + b = b + a:(1 + a)*(1 + b) = 1*(1 + b) + a*(1 + b)
= 1 + b + a + a*b(1 + a)*(1 + b) = (1 + a)*1 + (1 + a)*b = 1 +
a + b + a*bso a + b = b + a, and we have a division ring. Now,
by Wedderburn's Theorem, which is indeed nontrivial, a finite
division ring is a field.I would be surprised if any of these
axioms could be omitted, but to showthat none of them could,
you would have to construct 9 examples, each ofwhich satisfied
all hypotheses but one!If this is homework, then it is
hard!Derek Holt.===
===
Subject:
: Re: Induced completeness to
stronger metric spaces> Take a nonmpty set X and two metrics
d1(x,y) and d2(x,y) in X. Call> d2 stronger than d1 if:>
d1(x,y) <= C d2(x,y) forall x, y in X, where 1 <= C < oo> Let
d2 be stronger than d1.> It's easy to see that if (x_n) is a
Cauchy sequence in (X, d2) then> it's also Cauchy in (X, d1).
Now, assume (X, d1) is complete in the> sense that every
Cauchy sequence converges to some x in X. Does this> imply
that (X, d2) is also complete?Suppose E = {1/n : n = 1,2,3...}
and X = E U {0}. Let d1 be the euclidean metric on X. Define a
metric d2 on X by setting d2(x,y) = d1(x,y) if x and y are in
E, d2(0,x) = 1 = d2(x,0) for all x in E, and d2(0,0) = 0. Then
d2 is stronger than d1, but (x,d1) is complete while (X,d2) is
not.===
===
Subject:
: Re: Matrix math -- verify please?> explicitly
stated, so I wanted to verify:> - The product of two
rectangular matrices (e.g. A = 4x3 and B = 3x4) cannot> be
invertible, since each matrix has at most rank 3.> - The
product AB has at most rank 3. (But can be less)> - Even if A
has rank 3 and B has rank 3, AB may still have rank < 3. Oll
Korrect===
===
Subject:
: Re: I got low score on math test, please
advise me and take a lookA bit unclear what's happening. Would
a W be counted as units attempted? Ifyou met your deadline,
then the lack of proper results for successfullydropping and
deleting enrollment records for the course is merited. Your
truely bigger problem is learning the material and gaining
useful credit. That will require first, RESTUDY on your own,
and then COURSE REPETITION toactually study a second (or
third) time. Lack of concentrated, repeatedpractice and study
prevents some fairly normal people from succeeding
inMathematics. G C===
===
Subject:
: Re: Candy Inspiration (in the
news)> It would still be interesting to know, though, whether
there's any shapeof> ellipsoid whose *best* packing density
exceeds the *best* density for>
spheres.http://mathworld.wolfram.com/EllipsoidPacking.html--
Mark===
===
Subject:
: Re: Induced completeness to stronger metric
spaces>Suppose E = {1/n : n = 1,2,3...} and X = E U {0}. Let
d1 be the euclidean >metric on X.>Define a metric d2 on X by
setting d2(x,y) = d1(x,y) if x and >y are in E, d2(0,x) = 1 =
d2(x,0) for all x in E, and d2(0,0) = 0.>Then d2 is stronger
than d1, but (x,d1) is complete while (X,d2) is not.Which
Cauchy sequence in (X,d2) does not converge in (X,d2)?-- I'm
not interested in mathematics that might have anythingto do
with reality. -- Russell Easterly, in sci.math===
===
Subject:
: Re:
the anticlassicalist }{ i: linguistic negation> ... Has>
anybody tried to bring up a child as a native speaker?Yes.
http://www.wired.com/wired/archive/7.08/mustread.html?pg=8I'm
aware of other children who have been exposed to spoken
Klingonsince they were born, but it's been inconsistent (in
one case, thefather is the skilled speaker, and he's often
away on extendedmilitary service).> Now that would be>
interesting -- though they probably wouldn't want to advertise
the fact> to the local child welfare authorities. If Okrand
really did build some> counter-universal features into it, how
a child would deal with it as a> first language would be really
fun to watch.Language acquisition apparently works no matter
how un-natural thelanguage. The broken rules of Klingon seem
to be more a matter ofgoing against evolutionary pressures
rather than being contrary todeep human mental features. No
human language can do this is toostrong; the best we can
probably do is say That wouldn't survive inan evolving human
language.===
===
Subject:
: Re: question about periodic function> I
am having trouble empirically determining the period of the
following>function:What do you mean empirically? Why not do it
mathematically?>y = sin(2*pi*.018*t) + cos(2*pi*.02*t)>It is a
situation of superposition of two waves with a
slightly>differing frequencies (.018,.02). One can hear beats
in a tuning fork>mathematical sense it is strictly not a
periodic function with period>T as the definition f(x+T)=f(x)
is not satisfied.Nonsense. It is periodic with period T = 500,
because both the sin andthe cos components are.> But the
following>approximation is accepted in electronics engineering
wave form>analysis and acoustics.>Let 2 pi f t =w t be the
angular argument. >sin(w1 t) + cos (w2 t) = cos(pi/2- w1 t) +
cos (w2 t) >= 2 cos(pi/4+(w2-w1)t) cos(pi/4-w t) where w's
w1~w2~w are>approximately equal.>w2-w1 is neglected in
comparision to pi/4, making it sqrt(2)>cos(pi/4-w t).So it has
maximum value sqrt(2) at wt=pi/4 and regular>(high frequency)
time period 2 pi/w and beating time period 2>pi/(w1-w2)
appearing as the envelope in the compound function
graph.>Works out to 1/.019 and 1/.002 or 52.6 and 500 seconds
of regular(hf)>and beat periods respectively.No approximation
is needed.sin(2*pi*.018*t) + cos(2*pi*.02*t) = cos(2*pi*.02*t)
- cos(2*pi*(.018*t+.25)) =
-2*sin(2*pi*(.019*t+.125))*sin(2*pi*(.001*t-.125)) =
2*sin(2*pi*.019*(t-125))*sin(2*pi*.001*(t-125))So it's exactly
your familiar scenario of carrier and beats, butstarting at a
different time.===
===
Subject:
: Re: How many different resistances
with n resistors?We had a thread in rec.puzzles called 'R = PI
ohms' on this sort of topic awhile back. You'll still find it
in the archives (e.g. google groups) I===
===
Subject:
: Re: Can
religion explain the dinosaurs? by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
i1JIHht07455;><pre> The problem with all such (religious)
explanations> is that THERE IS NO GOD... and it's just some
con man> (or more usually, some madman) standing behind the>
curtain cutting off all questions as to why God/Allah> would
command such idiocy/evil with the impossible to> question:
Because God said so: I was there & I heard Him.> You may want
to temper your flaming atheism and admit that yours> is nothing
more than an opinion... you have less proof that God> doesn't
exist than many people do that He does.> Absurd. There is
zero proof that God doesn't exist for the same> reason that>
there is zero proof that anything in particular doesn't exist.
Try> proving> a negative. Any negative. I dare you. And if you
can't, why would you> require that of your opponents?> Sorry
to contradict but I have to lend support to Sebastian on>
this. He is absolutely right; there is more proof - or rather>
evidence - indicating the existence of God than atheists have>
for his nonxistence. It's no use waving that old chestnut>
unable to prove a negative about; the fact is that the small>
amount of evidence *for* negates any counter arguments> based
on nothing better than an inability to prove a negative.>
There is precisely zero proof for the existence or
nonxistence> of god.> If you have any specific proofs show
your stuff.>Sorry but you are wrong. To say zero proof is
quite a claim ->and he who makes claims must accept the burden
of proof so,>like Sebastian, I would caution you to be a little
careful with your>claims.>Note: I reduced my own claim above to
a small amount of>evidence - not proof. Of course the existence
of God is not>provable so I am not about to show you anything.
But evidence>is a quite different matter; qualitively it is
small indeed, but>quantitively it is massive, ranging from
(ostensible) miracles,>answered prayers, faith healing,
so-called paranormal>experiences, visions, ecstatic and
mystical experiences etc.,>etc. But I am sure you know all
this.>Whether or not you accept it as evidence is beside the
point ->it remains far more than you can offer as counter
argument.>-->altheim></pre>===
===
Subject:
: Re: Axioms defining a
finite field===>
===
Subject:
: Axioms defining a finite field> Let
(F, +, *) be a finite set with two operations> and constants
0, 1 such that the following rules hold:> (1) a + (b + c) =
(a + b) + c> (2) a + 0 = a> (3) for every a there's a b so
that a + b = 0> (4) a*(b*c) = (a*b)*c> (5) a*1 = a> (6) 1
is distinct from 0> (7) a*(b + c) = a*b + a*c> (8) (a + b)*c
= a*c + b*c> (9) a*b = 0 => a=0 or b=0> Show that F is a
field. Can one of the rules be omitted> so that F still has
to be a field?> let a /= 0, b0 = 0, b1 = 1> A = { a bj | 0 <=
j <= |F| }> A| = |F| because> if a bj = a bk: a(bj - bk) = 0;
bj - bk = 0; bj = bk> As F is finite, 1 in A> which shows a has
right multiplicative inverse a_r.> This with a1 = a, shows F0
is a group under *.> Did you forget a+b = b+a, ab = ba?> I
don't know about the a+b = b+a, but if we add that then> ab =
ba follows: it's a theorem that every finite division> ring is
a field. (It's not quite trivial, as I recall. Possibly> it
actually is trivial and just didn't seem trivial to me> at the
time; that was an algebra class when I was an>
undergraduate...)No, not trivial . This is Wedderburn's
theorem; Proofs from the Book givesa nice (what else?) four
page proof (chapter 5) due to Witt (and mention 7or 8 more,
using quite different ideas). Witt begins by proving,
usinglinear algebra, that the centraliser of s , is of
dimension q^(n_s) , whereq is the characteristic of F (q.1=0)
and that n_s divides |F|. Then, heimbeds F in the roots of
unity in C, and conclude by a clever argument onthe cyclotomic
polynomials...> ************************> ===
===
Subject:
: Re:
Non-uniqueness of factorization>Can someone please give me
step by step details on how to ignore posts by>James Harris
and replies to his posts? I just don't have the willpower
to>not read his
bull.http://www.hyphenologist.co.uk/killfile/
killfilefaq.htm===
===
Subject:
: Re: strain softening spring>i.e.
the equation I am trying to solve is as follows:>y'' + cy' +
[ko*e^(-alpha*t)]y = 0>Any help would be most
appreciated.>sincerely>Paul Joseph>(pjoseph@excite.com)> Maple
gives a solution in terms of Bessel functions. I posted a file>
called DE.pdf showing the solution at:>
http://math.asu.edu/~kurtz/de/> --LynnHi Lynn,This is in
regard to the solution you kindly posted above.Can the first
argument of the BesselY function be negative? I triedto
implement this in Excel, but Excel's BesselY function appears
torequire that the first argument be positive. I am not sure
if this isa quirk of Excel... or whether this is a
mathematical requirement.Since c and a are both real and
positive numbers, -c/a will
benegative...no?thanks!Paul===
===
Subject:
: Model theory puzzleFind
a finite system of axioms (feel free to introduce
operations,relations, constants) so that all of its finite
models would have aprime number of elements, and for every
prime p there's a model tothat system of size p. Posted Via
Usenet.com Premium Usenet Newsgroup
Services------------------------------------------------------
---- ** SPEED ** RETENTION ** COMPLETION ** ANONYMITY
**----------------------------------------------------------
http://www.usenet.com===
===
Subject:
: Single Precision Montgomery
Multiplication OptimizationThe single precision Montgomery
multiplication algorithm can be stated asfollows:$$eqalign { t
&= x y,cr u &= (t m^prime) bmod b,cr v &= u m,cr A &= (t + v /
b) bmod m.cr}$$where $m^prime = - m^{-1} pmod b$ and $b$ is
the radix.It would seem that the term$$x y + (x y -m^{-1} bmod
b) m$$can be reduced in some way so that only two or even one
multiplication is===
===
Subject:
: Re: Peer Review
<TClTb.6252$jH6.3922@newsread1.news.atl.earthlink.net>
<8083d02b.0402031024.42779ed2@posting.google.com>
<bvoq6b$bi1$1@newsg3.svr.pol.co.uk>
<8083d02b.0402181638.267a31f8@posting.google.com> Discussion,
linux)> Another anecdote:> Simon Singh in his book Fermat's
Last Theorem notes what may have> been the first instance of
peer review injustice. Hippasus in the> school of Pythagoras
realized that irrational numbers exist. This was> regarded as
heresy and he was put to death for it.But this story is almost
certainly false, isn't it?-- Even if [...] a communistic regime
should come [to China], the oldtradition [...] will break
Communism and change it beyond recognition,rather than
Communism [...] break the old tradition. It must be so. -- Lin
Yutang on Socialism with Chinese characteristics in
1935===
===
Subject:
: Re: countable sets> Let E1, E2, E3, ... be a
sequence of pairwise disjoint countable sets. Prove> that
union of Ek as k = 1 to oo is countable. I do not need help
proving> this. I need someone to explain exactly what is
being asked. In Let E1, E2,> E3,... be a sequence of pairwise
disjoint countable sets is that saying> that the numbers of
Ei's are countable or is it saying that each Ei has a>
countable number of elements??> Others have explained what is
being asked, but I was intrigued by your> statement that you
don't need help proving it. Now that the question has> been
clarified, you might look again at proving it.> Hint: If your
proof does not invoke the axiom of choice, then your proof> is
wrong.>Not so. The countable union of disjoint countable
sets>is in 1-1 correspondence with NxN -- the pair (n, k) maps
to>the the k-th element of the n-th set (the set of sets and
each>set is alreadyknown to be in 1-1- correspodence with N).
This>is clearly surjctive, and the disjointness implies that
it's>injective.This part involves some of the Axiom of Choice.
Solovayhas a model in which the reals are a countable union
ofcountable sets. Each of the sets can be enumerated, butnot
all at once. >And it does not require AC to show that NxN is
countable.If one has simultaneous enumeration, then it is
reducedto this, which is as you say.-- This address is for
information only. I do not claim that these viewsare those of
the Statistics Department or of Purdue University.Herman
Rubin, Department of Statistics, Purdue
Universityhrubin@stat.purdue.edu Phone: (765)494-6054 FAX:
(765)494-0558===
===
Subject:
: Re: I got low score on math test,
please advise me and take a look> my website states my case
and has jpg files of the four pages of the> test please take a
look and advise me or give me opinions>
http://www.johncho.usAside from the all the politics of
whether the grade was fair and if thetest should be returned
in time, I have a few comments. First, for all the instructors
out there, how many of you take of 1/4 ofa point? To me, if you
make the problem worth 8 points, anything lessthan perfection
is 7. And, when I am assigning point values toproblems, I
usually look over the problem and pick out the two or
threeconcepts being tested in that question and give each of
those concepts avalue of 1 or 2 or even 3 points. I just can't
igaine trying to gradewith fractional points. But, that is just
my grading style and I'm notreally saying anything is wrong
with giving fractional points,I justfind it uncommon.
Secondly, if I am grading and I cannot follow the work
(perhaps becauseit is not neatly written), I have a hard time
giving full credit. Thatmight sound harsh, but I have often
found that students write two orthree possible answers for the
problem in their mess and never clearlymark which one they want
me to grade. Also, judging from your scans ofthe page, it looks
like you had the back of each page left blank (whichis pure
speculation on my part). I despise it when students try
tosqueeze in something in the margins, leaving the whole back
side of apage blank. You aren't going to use that page for
anything else, whysqueeze things together? Now, if you really
want to get your instructormad, you could ask him (or her) to
please write the comments on yourtest neater. But that would
serve no purpose otehr than to point youout as a snot-nosed
brat. Okay, about the grading. Personally, I don't know what
topics werestressed in class so I don't feel like I can make a
judgement call. Forexample, I gave a test in Trigonometry II on
Tuesday (which I am in theprocess of grading and won't be given
back until Monday). For this testthe material I stressed was
changing degrees to radians, law of sines,and law of cosines.
On one question I asked about the area of atriangle. If a
student accidentally used the formula area = base *height, I
will not take off many points probably one out of 10
points,because they were not being tested on geometry (and had
they asked methe correct formula, I would have told them, since
that wasn't the topicthey were being tested on). Or, maybe
another example, if someonecomputed 2+3=6 in part of a larger
problem, multiplying instead ofadding, and they correctly
carried that mistake through the problem(that is if they had
used 5 they would have gotten the correct answer) Iwould
probably only take off one point out of 10. Those are my
thoughts. Your grade, suck it up. If you ask theinstructor to
look over the test again, he or she might find some placeswhen
you should hae lost more points that you did. At least, if
someonequibles over two or three points I say I will look it
over, and I do andmake sure I *took off* all the points I
should have. - Tim-- Timothy M. BrauchGraduate
StudentDepartment of MathematicsWake Forest Universityemail
is:news (dot) post (at) tbrauch (dot) com===
===
Subject:
: Re:
Solving Equation with lnHumm, I am not familier with the
Lambert W function, is there a website you would recommend so
that I can read up on it or is thisfunction too complex for
the non-math major type person. Andrew V. RomeroI was working
on a math problem at work and ran into something I don'tknow
how to solve and was hoping someone could give me a clue on
thisfairly simple problem.Original Equation:227.11 =
5.8{ln(38/x)} + (38-x)Simplfied Equation:189.11 =
5.8*{ln(38/x)} - xof course you can reduce this down to32.60 =
ln(38/x) - x/5.8and further (I think) to 1.44 x 10^14 = 3.8/x -
e^(x/5.8)but I am stuck after this. Is it possible to solve for
X?> Not in terms of the standard elementary functions. But you>
can solve in terms of the Lambert W function...> x =
W(38*exp(-(189.11)/5.8)/5.8)*5.8 =
2.627466197*10^(-13);===
===
Subject:
: Re: HELL :-) (Was: Re: the
anticlassicalist }{ ii: the spectre continues)[...]mitch, I
don't snip here as a skip. I just wanted to say that I see
clearlyyou are one who has already learned that the only
secret behind any ofthis is the hunt for patterns. Going out
and grabbing disparate works tolook for connections, then
hunting down connections to those, and buildingthem into
pictures. When the connections become rigorously symbolised,
wehave math (or logic, or numerology, or symbology -- whatever
you like).When a model connects ontology to epistemology, a
theory is born. Its nice watching people have fun with their
patterns, play with them,fit the pieces together. Its almost
as fun as doing it myself. I wanted to say this because I see
many on usenet who frown on suchbehavior. They seem to look
only for some kind of stale acceptance of anever surprising
universe, and look to attack any excitement or creativitythey
see. So I sometimes want to balance the powers, give praise at
theinteresting twists and turns I see people playing with. I
have a child likewonder for child like wonder. You even picked
up Conway and Sloan... That was the first book I everchecked
out of a university library (I didn't understand it then, but
I keptcoming back to try again).--
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-galathaea:
prankster, fablist, magician, liar===
===
Subject:
: Re: Model theory
puzzle> Find a finite system of axioms (feel free to introduce
operations,> relations, constants) so that all of its finite
models would have a> prime number of elements, and for every
prime p there's a model to> that system of size p. Go
ahead!===
===
Subject:
: Re: covering compact set w/squares>Oh, I
think I did misread. How about this:>Given a compact set K in
the plane s.t. each pt x is the center of a square>Q_x, prove
that you can find a subsequence Q_x_i of squares s.t. K
is>covered by the Q_x_i and>sum(over i) Char(Q_x_i) <= 4 Char
(union(over i) Q_x_i),>where Char(X) is the characteristic
(indicator) function of X.Are you assuming the squares are all
oriented with sides parallel to the axes? If not, there's a
counterexample with K consisting of 7 points at the vertices
of a regular heptagon; take a square centered at each ofthese
points, oriented so the centre of the heptagon is on a
diagonalof the square, and large enough so that the square
contains aneighbourhood of the centre of the heptagon but each
point of K isoutside the squares centred at other points of K.
It makes arather pretty picture:
<http://www.math.ubc.ca/~israel/sevensquares.gif>.===
===
Subject:
:
Re: RFI:Wanless' Fourth Conjecture/Dirichlet's Geometric
TheoremSTOP PRESS!! :-((I've been incredibly stupid!The
generalized form is clearly a load of dingo's kidneys: -take
a=5, b=3 for example - a^n+b will _always_ be even!!Don't
waste any more time on it...J> The series a^n+b [hcf(a,b)=1]
includes an infinite number of primes [over> all n]===
===
Subject:
:
Re: strain softening spring>Can the first argument of the
BesselY function be negative? I tried>to implement this in
Excel, but Excel's BesselY function appears to>require that
the first argument be positive. I am not sure if this is>a
quirk of Excel... or whether this is a mathematical
requirement.>Since c and a are both real and positive numbers,
-c/a will be>negative...no?>thanks!>PaulApparently this is a
quirk of Excel (you did notice that Excelrequires the
arguments in the reverse order?). You can read more
aboutproperties of BesselY
athttp://functions.wolfram.com/BesselAiryStruveFunctions/
BesselY/If you have particular values of a, c, and k, I could
plug them in andpost a graph of the solution for you. Probably
Maple or Mathematicawould be a more appropriate tool for
you.--Lynn===
===
Subject:
: Re: Are the derivatives of abs[(x-a)^3]
different for x>a and x<a ?> given abs[(x-a)^3], where abs[ ]
means taking the absolute value, and> a is a constant, are the
first, second, third derivatives (with> respect to x) different
for x>a and x<a ?> I forgot how to write down the derivative of
an absolute function in a> formal way, I mean, not using if
else, but using something likesign(x-a).> I think the first
derivative would look like:> 3(x-a)^2 * sign(x-a)Take a = 0
for convenience. Let f(x) = |x|^3. Then f(x) = x^3, x >= 0,
f(x) = -x^3, x <= 0. So f'(x) = 3x^2 for x >= 0, f'(x) =
-3x^2, x <= 0, f''(x) = 6x for x >= 0, f''(x) = -6x for x <=
0. And f'''(0) doesn't exist. You can of course use the sgn
function to express these if you like. All the claims about
derivatives at 0 need to be checked using the basic definition
of a derivative.===
===
Subject:
: Re: Are the derivatives of
abs[(x-a)^3] different for x>a and x<a ?> The usual method is
to define sgn(x) in terms of the heaviside function> H(x) = {
0, x < 0; 1, x > 0; 0.5, x = 0 }, and write formally> dH/dx =
delta(x) where delta(x) = { 0, x != 0 } is the Dirac delta>
'function' (integral of delta(x) over any interval containing
0 is 1 by> definition, and 1/2 if 0 is an endpoint).This is a
joke, right?===
===
Subject:
: Re: the anticlassicalist }{ i:
linguistic negation... Hasanybody tried to bring up a child as
a native speaker?> Yes.
http://www.wired.com/wired/archive/7.08/mustread.html?pg=8>
I'm aware of other children who have been exposed to spoken
Klingon> since they were born, but it's been inconsistent (in
one case, the> father is the skilled speaker, and he's often
away on extended> military service).Now that would
beinteresting -- though they probably wouldn't want to
advertise the factto the local child welfare authorities. If
Okrand really did build somecounter-universal features into
it, how a child would deal with it as afirst language would be
really fun to watch.> Language acquisition apparently works no
matter how un-natural the> language. The broken rules of
Klingon seem to be more a matter of> going against
evolutionary pressures rather than being contrary to> deep
human mental features. No human language can do this is too>
strong; the best we can probably do is say That wouldn't
survive in> an evolving human language.According to a strict
Chomskyan view, a language constructed contrary toUG should be
un-learnable (i.e. un-acquirable in the normalfirst-language
way, though of course you could learn it as anintellectual
exercise). Presumably the child would re-structure it
intosomething else which conformed with UG. However, from the
examplesgiven, it appears as Jacques says that the
counter-universal featuresare rather trivial or perhaps not
counter-universal at all.Ross Clark===
===
Subject:
: Re: Coprime
Grid: Filling Infinite Quadrantath: meganewsservers.com>In
this post, I write of a specific and altered case of the
puzzle>mentioned in these previous threads:http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&safe=off&
threadm=b4be2fdf.0211011659.79913415%40posting.google.com&rnum
=3&prev=http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&safe=off&
threadm=b4be2fdf.0401241437.73a12678%40posting.google.com&rnum
=4&prev=http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&safe=off&
threadm=b4be2fdf.0303071814.63262520%40posting.google.com&rnum
=1&prev=>In the puzzle, we try to write, in order, the
positive integers into a>grid, one integer per grid-square,
such that:>Each integer n is adjacent (above /left of /right
of /below)>of the integer (n+1);>And each adjacent pair of
integers (above /left of /right of /below)>are coprime.>And
the goal is to completely fill the grid.>But here I am
asking about filling an infinite grid which is bounded>along 2
perpendicular sides.>ie. the grid is an entire quadrant of the
Cartesian plane, bounded by>the x-axis and the y-axis.>I think
I found a simple procedure which *might* ensure a
successful>filling of the grid with coprime-adjacent
integers.>(Sorry to those on rec.puzzles, but I will give my
procedure below.>You can still post your own algorithm,
however, or confirm that mine>can really work for the entire
grid without problems.)>I illustrate with the first 99
terms:>(figured by hand, so maybe inrror)>99>98>97 96 95>54
55 94 93 92>53 56 57 58 91 90 89>52 51 50 59 60 61 88 87>21 22
49 48 47 62 63 86>20 23 24 25 46 45 64 85>19 18 17 26 27 44 65
84>06 07 16 15 28 43 66 83 82 81 80 79>05 08 09 14 29 42 67 68
69 70 71 78>04 03 10 13 30 41 40 39 38 37 72 77 76>01 02 11 12
31 32 33 34 35 36 73 74 75>Basically, the path swings
clockwise and counterclockwise, running>along the outside of
the already-filled section.>When it gets to either the x-axis
or y-axis, it forms a 'peninsula',>the length of which is the
shortest needed to avoid uncoprime integers>being placed next
to each other in the path-section which runs from>that
peninsula's axis to the other axis.>Now, we do not want a
situation where, following the algorith>precisely,>there is NO
peninsula-length which would avoid uncoprime neighbors.>I am
not certain, but I believe this issue is not a problem.>Fun
perhaps: Show if my algorithm is foolproof...or just
foolish.> Well, what I know about number theory and 50
cents will get me a cup> of coffee, but it seems to me that
the way to attack this solution> is to show that once the
peninsula length exceeds some number N,> you will always have
a noncoprime pair when it loops its way back.> It seems to
me that this ought to be true. > Of course, that stil
wouldn't prove that your algorithm can't> work, because you
might never have to buld a peninsula out that big,> but it
would be a start.>(I know I do not give the peninsula's
length. Also fun perhaps: try to>determine the shortest length
needed for each peninsula, given that>the algorithm never leads
to a problem.)>Leroy Quet> George>I do not believe you are
necessarily right about the filling-algorithm>working for all
peninsula-lengths above N(k), for the k_th peninsula.That's
not what I said. I'm not trying to make your algorithm
work,I'm trying to show it won't work, I find that to be more
fun :-)My thought was that, in general, the longer, the
peninsula,the more likely there will be at least 1 pair of
noncoprimenumbers when it loops back. So maybe beyond a
certain lengthyou will always have at least one such pair.For
example, if I start at one, and try various peninsula
lengths,2 and 3 are ok, but 4 doesn't work because 3 matches
with 61 2 3 48 7 6 5further, any penisula with lenngth 3n+1
will also have a problem withmultiples of 3 matching1 2 3 4 5
6 714 13 12 11 10 9 8similarly, if the peninsula is of length
7 then multiples of 5 willmatch, and there will be a match for
any peninsula length 5n+2.similarly, for a peninsula of length
10 then 7 will mathch 14, andyou'll get a match for every
peninsula lentgh of 7n+ 3So the density of possible peninsula
lengths decreases in a manneranalogous to the way that the
density of primes themselves.But although there are an
infinite number of primes, it's not clear tome right now if
there are an infinite number of possible peninsulalengths. To
those of you who know lots of number theory,it may be really
obvious, but it isn't to me.George>Leroy Quet===
===
Subject:
: Re:
Model theory puzzle> Find a finite system of axioms (feel free
to introduce operations,> relations, constants) so that all of
its finite models would have a> prime number of elements, and
for every prime p there's a model to> that system of size
p.Hint: forget about model theory as such and ask yourself if
you canthink of an algebraic structure that always has a prime
number ofelements (or a particular kind of well-known algebraic
structure...).Hthab-- G.C.===
===
Subject:
: Re: Induced completeness
to stronger metric spaces>Suppose E = {1/n : n = 1,2,3...} and
X = E U {0}. Let d1 be the euclidean >metric on X.>Define a
metric d2 on X by setting d2(x,y) = d1(x,y) if x and >y are in
E, d2(0,x) = 1 = d2(x,0) for all x in E, and d2(0,0) = 0.>Then
d2 is stronger than d1, but (x,d1) is complete while (X,d2) is
not.> Which Cauchy sequence in (X,d2) does not converge in
(X,d2)?Note that the sequence S = {1, 1/2, 1/3, 1/4, ..., 1/n,
...}converges in (X,d1), therefore must be Cauchy in
(X,d1).Note further, that the distances among the elements of
S are thesame wrt d2 as wrt d1.Conclude that (since the
definition of Cauchy sequence is in termsof interlement
distances only) S is a Cauchy sequence in (X,d2).Although one
might prefer S to converge to 0, it cannot do so in
(X,d2),since all distances from S to 0 are 1.Wasn't that kind
of obvious?Dale.===
===
Subject:
: Re: No Set Contains Every
Computable Natural> Given any set, S, of representations of
natural numbers,> I can write a TM that will find a
representation of a natural number> that is not in S. (No,
this TM does not halt after a finite number> of steps.)This
is, of course, simply false, false, false.If you give the set
of all representations of natural numbers, thenyour TM will
not find any representation of a natural number not inyour
set.You're speaking nonsense again.Of course, we must be
careful here and fix a convention for when atape represents a
set of natural numbers. Let's say that a taperepresents a set
of natural numbers if it consists of blocks of onesseparated
by a single space.We must also be careful to require that your
TM has the property that,for each square x of the tape, there
exists a steps t such that for allsteps s after t, the symbol
on the square x at time s is the same asthe symbol on x at t.
After all, a TM which repeatedly changes the0th square to 1
and blank and 1 and blank cannot be said to produceany tape at
all.> I have given this proof several times.> My TM has the
property you describe.Now, on an input tape containing 1,
,1,1, ,1,1,1, ,1,1,1,1, ,...,> How about
b1b11b111b1111...there is simply no possibility that your TM,
even after an infinitenumber of steps, creates a tape with a
representation of some naturalnumber not on my input tape.
Every natural number is represented onmy tape. None is
missing. Your tape either returns a set of numbersalready on
my tape or it returns no set of numbers at all.> This TM will
find a representation not on your tape:> It is a three state
machine and I can provide a state> transition table if you
like.> 1) Find a blank> 2) Find a second blank> 3) Backup and
write a 1 on the previous blank> repeat steps (1) through (3)>
This TM will produce a tape that contains exactly one blank.>
The contiguous string of 1's preceding this blank will be> a
representation not on your tape.> (Others have noted this is a
very crude adder.)Thus, when the TM reaches 'n' on the tape, it
will produce the number 'n*(n+1)/2+n-1', which is still further
along on the tape (but already on the tape).What number 'n'
does your TM ever reach for which 'n*(n+1)/2+n-1' is not
already on the tape?Answer: there is no such 'n'.> Obviously,
a human has to decide if a symbol represents> a natural
number. No TM can do this.> No TM can devise a convention, if
that's what you mean. At least, no> TM can devise a convention
in the sense that matters here.> But, a TM can extend the
convention in such a way that no set> contains every symbol
that represents a natural number.No, no, no. A TM cannot
extend the representation of natural numbersfor two reasons:
(1) It's a ing Turing machine, isn't it? I don'twant to
argue about requirements for intentionality or whether TMs
arecapable of intelligence, but this TM is not part of the
negotiationsregarding our conventions.> No intelligence
required.> My TM just adds all the numbers> together (plus 1
for each addition).> A TM is too stupid to know that> the sum
is supposed to be infinity.(2) Every natural number has
arepresentation in our convention, so the representation
cannot benon-redundantly extended.> I just described such a
TM.> No tape can contain every representation of a natural
number.> There will always be a bigger representation.No
finite tape, but as soon as you posit a tape containing a
representation for every natural, you have such a tape as does
not allow any bigger representations.> Russell> - 2 many 2
count===
===
Subject:
: Re: How many ways to put 5 balls into 500
ordered cups?> SUMMARY: (1) Find the ways to partition the
number 5: 5 4 1 3 2 3 1 1 2 2 1 2 1 1 1 1 1 1 1 1 <<The
gimmick is that the cups are ordered, so (2,3) is not the same
as(3,2) fo the two cup case and so forth upt to 500.===
===
Subject:
:
Re: . The hardest of all hard facts .Hi Eleaticus, Re: How you
think, Michelson-Morley and Kennedy-Thorndike do indeed fit
Galilean ( c + v ) physics ,All throughout the annals of
history ... No premise has been better tested than this
premise: The speed of light is the same for all observers.That
makes it: The hardest of all hard facts.> No experiment ever
showed that the speed of light is> the same for all observers.
Indeed any determination> of the one way speed of light can be
used to demonstrate> the speed of light is NOT the same for
all observers....> A light----> B <-you> < ----------- L
--------------> v m/s> Use syncronised clocks at A amd B to
time how long it takes> light to travel a distance of L meters
across the laboratory..> Speed of light relative to the
laboratory = L/ (tB - tA) = c> where 'tA' is the time at which
the light left A> and 'tB' is the at which the light arrived at
B> Now repeat the experiment while running towards B at v m/s>
Note that 'in your frame of reference' the point B is moving
,> so that the light must travel an extra distance = v * (tB -
tA)> which is the distance B has moved as the light travels
from> A to B.> Therefore:> Speed of light relative to you =
(L+ v * (tB - tA)) / (tB - tA)> = c + v> keith steinYou had
measured the relative velocity between a light ray front and
amoving you with v velocity. Notice that the velocities of
your twomoving entities are c and v, measured in the
Laboratory inertialframe. As c and v refer to the same
inertial frame, you have the rightto use ordinary vector
algebra obtaining c+v, because Relativevelocity can be >c and
is NOT invariant (see my post with thattitle). If you measured
the relative velocity in the inertial frameyou is at rest you
obtain c as the result, compatible with thesecond postulate of
Special Relativity. ANY light continue having thesame vacuum
speed c for ANY observer AT REST in ANY inertial frame,The
hardest of all hard facts .RVHG===
===
Subject:
: Re: . The hardest
of all hard facts .Hi Eleaticus, Re: How you think,
Michelson-Morley and Kennedy-Thorndike do indeed fit Galilean
( c + v ) physics ,All throughout the annals of history ... No
premise has been better tested than this premise: The speed of
light is the same for all observers.That makes it: The hardest
of all hard facts.> No experiment ever showed that the speed of
light is> the same for all observers. Indeed any determination>
of the one way speed of light can be used to demonstrate> the
speed of light is NOT the same for all observers....> A
light----> B <-you> < ----------- L --------------> v m/s> Use
syncronised clocks at A amd B to time how long it takes> light
to travel a distance of L meters across the laboratory..>
Speed of light relative to the laboratory = L/ (tB - tA) = c>
where 'tA' is the time at which the light left A> and 'tB' is
the at which the light arrived at B> Now repeat the experiment
while running towards B at v m/s> Note that 'in your frame of
reference' the point B is moving ,> so that the light must
travel an extra distance = v * (tB - tA)> which is the
distance B has moved as the light travels from> A to B.>
Therefore:> Speed of light relative to you = (L+ v * (tB -
tA)) / (tB - tA)> = c + v> keith steinYou had measured the
relative velocity between a light ray front and amoving you
with v velocity. Notice that the velocities of your twomoving
entities are c and v, measured in the Laboratory
inertialframe. As c and v refer to the same inertial frame,
you have the rightto use ordinary vector algebra obtaining
c+v, because Relativevelocity can be >c and is NOT invariant
(see my post with thattitle). If you measured the relative
velocity in the inertial frameyou is at rest you obtain c as
the result, compatible with thesecond postulate of Special
Relativity. ANY light continue having thesame vacuum speed c
for ANY observer AT REST in ANY inertial frame,The hardest of
all hard facts .RVHG===
===
Subject:
: Re: JSH: Splitting field,
algebraic integer factors[snipped unneeded restatement of
errors already claimed and reclaimed too many times]> Well I
was wrong. I should have known in retrospect, but it was yet>
another situation where I kind of decided that I wanted
something and> thought I had it, only to find out later I was
wrong.> There's no point to hanging on to wrong things
though.> Luckily, mathematics is supposed to be an arena where
the truth> matters.> James HarrisAnd, unluckily for JSH's
egomania, the truth will out.===
===
Subject:
: Re: I got low score
on math test, please advise me and take a lookA 77 is not that
bad in a begining Calc class. Why would you want todrop?Just
study more and practice more problems for the next test, you
canbring the grade up.Dropping a class is going to make your
college years slower, becauseyou are going to have to take
this class sometimes , and who knowsyour next teacher might be
harder.Plus, write more legible on the next test.===
===
Subject:
:
Re: Definition of Separable Space (basic topology
question)>Let A be a subset of (X,T). Then A is dense in X iff
for every nonmpty>open subset U of X, A / U != {}.>I have seen
two definitions of 'separable topological space':>a) (X,T) is
separable if there exists A <subset> X, where A is countable
and>dense in X>b) (X,T) is separable if there exists A <proper
subset> X, where A is>countable and dense in X>Given def (a),
its simple to prove that all countable spaces X
are>automatically separable since the subset X is countable
and non-trivially>intersects every nonmpty open subset.
However, this won't work given def>(b). I first became
suspicious when I saw a proof that all countable spaces>were
separable that seemed ridiculously complex in comparison to
the>seemingly obvious 1 step proof above. I later found
definitions of>separable like def (b).The integers with the
usual topology is separable. Definition (a) is the correct
one. BTW, finite spaces are separable.-- This address is for
information only. I do not claim that these viewsare those of
the Statistics Department or of Purdue University.Herman
Rubin, Department of Statistics, Purdue
Universityhrubin@stat.purdue.edu Phone: (765)494-6054 FAX:
(765)494-0558===
===
Subject:
: Re: I got low score on math test,
please advise me and take a lookX-ID:
buRBAoZZ8ep-0neGpdIfA21ofva91P3Udt3Zut1ttQkdUU8eCml20c David
C. Ullrich <ullrich@math.okstate.edu> schrieb:>problem 9: 4.5
out of 5. I would have given you fewer points>than that,
because there's a bit of explanation missing>(you need to
point out that the function is not even>_defined_ at the point
in question, because that>denominator vanishes).Am I missing
something here? How can you say that a function
isdiscontinuous at a point, where it's not defined? That
function iscontinuous at each point of its domain - Unless of
course you definef(2) = 3.01 or something like
that.===
===
Subject:
: Re: smallest eigenvalue of Laplacian> there
is a theorem that says that an eigenfunction corresponding to>
the smallest eigenvalue of the Laplacian on some bounded domain
has no> zeros inside that domain.> Is the reverse implication
also true? Meaning: Does every> eigenfunction without zeros
inside the domain belong to the smallest> eigenvalue?> Thank
you in advance for any helpful comments.> Yours sincerely,>
Tobias N.8ahringWhat a great question. Someone told me that it
has been recently shown that in two dimensions, the
eigenfunction corresponding to the second eigenvalue has
zeroes, indeed, the set of zeroes forms a nodal line that
splits the domain in half. Presumably if you look at this
paper, it might have references to the result you are looking
for. I think that this is the paper:MR1152231
(93g:35100)Melas, Antonios D.(1-UCLA)On the nodal line of the
second eigenfunction of the Laplacian in $ Rsp 2$.J.
Differential Geom. 35 (1992), no. 1, 255--263.35P05 (35J05
58G25) eigenfunction of the Laplacian with zero boundary
condition for a bounded domain $Omegasubseteqbold R^2$ does
not have a closed nodal line. This was asked by S.-T. Yau for
$Omega$ a bounded convex domain in $bold R^2$. Twenty years
ago, Payne proved the conjecture provided the domain
$Omegasubseteqbold R^2$ is symmetric with respect to one line
and convex with respect to the direction vertical to this
line. Also, C.-S. Lin proved the conjecture provided the
domain $Omegasubseteqbold R^2$ is smooth, convex, and
invariant under a rotation with angle $2pi p/q$, where $p$ and
$q$ are positive integers. Recently D. Jerison [Internat. Math.
Res. Notices 1991, no. 1, 1--5; MR 92d:35210] proved the
conjecture for long thin convex sets. Without any assumption
on the smoothness of $Omega$ he showed that the nodal line has
to intersect $partialOmega$ in exactly two points.We prove the
conjecture when $Omega$ is a bounded convex domain in $bold
R^2$ with $C^infty$ boundary.===
===
Subject:
: Re: x^2 + y^4 =
z^4===>
===
Subject:
: Re: x^2 + y^4 = z^4> the equation x^2 +
y^4 = z^4 has no positive integer solutions.> Is the proof
... short enough for some kind soul to post it>An elementary
proof is based on descent. That is, assume a>solution exists
in positive x, y, z, and show that another one can be>found
with smaller z.> Are you sure?> That's true of x^4 + y^4 = z^2
where infinite descent is used at>
http://www.math.toronto.edu/mathnet/plain/questionCorner/
fermat4.html> However I see no way to modify the method used
there to this problem.> ----Here is an elementary proof by
descent.We may assume WLOG that x, y and z are pairwise
coprime, and thatthere is a minimal solution with xyz <> 0.
There are two cases:I. x odd, y even, z oddAssume that y is
minimal and nonzero.x = a^2 - b^2y^2 = 2abz^2 = a^2 + b^2 with
a, b coprime and of opposite parity [known parametric
solution]a = c^2b = 2d^2 c, d coprime, c odd (alternatively a
= 2c^2, b = d^2, which leads to the same result) [follows from
y^2 = 2ab with gcd(a,b) = 1]a = e^2 - f^2b = 2ef e, f coprime
and of opposite parity [follows from z^2 = a^2 + b^2 with a
odd, b even, a coprime to b]d^2 = efc^2 = e^2 - f^2 [combining
the above]e = g^2f = h^2 [follows from d^2 = ef with gcd(e,f) =
1]c^2 = g^4 - h^4c^2 + h^4 = g^4 [by substitution]y^2 = 2ab =
4c^2 d^2 = 4ef(e^2-f^2) = 4g^2 h^2 (g^4-h^4) > h^2 > 0 [by
substitution and assuming y <> 0]Thus c^2 + h^4 = g^4 is a
solution with 0 < |h| < |y|, whichcontradicts the assumption
that y is minimal and nonzero. QED case I.II. x even, y odd, z
oddAssume that x is minimal and nonzero.x = 2aby^2 = a^2 -
b^2z^2 = a^2 + b^2 with a, b coprime, a odd, b even [known
parametric solution]a = c^2 - d^2b = 2cd c, d coprime and of
opposite parity [follows from z^2 = a^2 + b^2]a = e^2 + f^2b =
2ef e, f coprime and of opposite parity [follows from y^2 = a^2
- b^2]cd = efc^2 - d^2 = e^2 + f^2 c, e odd, d, f even
[combining the above]c = ghd = ije = gjf = hi g, h, i, j
pairwise coprime g, h, j odd, i even [follows from cd = ef,
gcd(c,d) = gcd(e,f) = 1, g = gcd(c,e), etc.]g^2 h^2 - i^2 j^2
= g^2 j^2 + h^2 i^2 [by substitution]g^2 (h^2 - j^2) = i^2
(h^2 + j^2) [by rearrangement]2g^2 = h^2 + j^22i^2 = h^2 - j^2
[follows from gcd(g,i) = gcd(h,j) = 1, g, h, j odd, i even]h^2
= g^2 + i^2j^2 = g^2 - i^2 [by addition and subtraction]k =
2gik^2 + j^4 = h^4 [by substitution]x = 2ab = 4cd(c^2-d^2) =
4ghij(g^2 h^2-i^2 j^2) = 2hjk(g^2 h^2-i^2j^2) > k > 0 [by
substitution and assuming x <> 0]Thus k^2 + j^4 = h^4 is a
solution with 0 < |k| < |x|, whichcontradicts the assumption
that x is minimal and nonzero. QED case II.===
===
Subject:
: Re:
Nonlinear PDE Help>Here's a partial differential equation that
I'm almost positive>has a unique solution, although I don't
have any idea how to go>about finding it. Does anyone have
suggestions?>We want to find f(x,y) subject to the following
two conditions> 1. f(x,0) = g(x)> 2. (df/dx)^2 + (df/dy)^2 =
1>where g(x) is some known differentiable function of x>and
where d/dx and d/dy are supposed to be partial derivatives.>I
know of a solution for a specific case: g(x) = square-root(1 +
x^2).>Then a solution is> f(x,y) = square-root(x^2 +
(y+1)^2)(if you're not including the point [0,-1] in the
region you're interested in)Geometrically, your differential
equation says that gradient(f) has length 1 everywhere. In
principle, you should be able to get solutions bystarting with
a more-or-less arbitrary curve C (either closed, or going to
infinity at both ends), on which you take f = c for some
constant c, and defining f(x,y) = c + dist((x,y), C) on one
side of the curve, = c - dist((x,y), C) on the other side of
the curvewhere dist((x,y), C) is the minimum distance from
(x,y) to C. Of course,depending on the curve, this may be
non-differentiable at some points(which have more than one
closest point on C). Two easy special casesare where C is a
circle (as in your example) or a straight line.By the way, the
latter example shows that in general f(x,0) = g(x) does not
uniquely determine the solution: consider f(x,y) = y and
f(x,y) = -y, both with g(x) = 0.I don't know if there are
other solutions (besides circles and straight lines) with nice
simple formulas: for most curves, calculating the distance from
a point to the curve is not very easy.===
===
Subject:
: Re: smallest
eigenvalue of Laplacian>there is a theorem that says that an
eigenfunction corresponding to>the smallest eigenvalue of the
Laplacian on some bounded domain has no>zeros inside that
domain.>Is the reverse implication also true? Meaning: Does
every>eigenfunction without zeros inside the domain belong to
the smallest>eigenvalue?The Laplacian is a self-adjoint
operator, so eigenfunctions for differenteigenvalues are
orthogonal. Two functions that both have constant signscan't
be orthogonal, unless one is 0.===
===
Subject:
: Re: JSH:
Non-uniqueness of factorization by support1.mathforum.org
(8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id
i1JKcdt19789;>That's why I hate posters like Dik Winter and
Nora Baron because>they're so adept at hiding things.>Like
notice how they used the example of>x(x+1)/2>always being an
integer in the ring of integers?>That's because the ring of
integers is a unique factorization domain,>so you can factor
*every* integer as 2j(j+1) or j(2j+1) with j an>integer in the
ring of integers.just curious - how would you factor 1, 5, 6,
7, 8, or 9in the form 2j(j + 1) or j(2j + 1) ??? and a few
others.>Integers are a unique factorization domain!!!yup.>Now
how many of you know that but listened to those posters?I
dunno. It's hard to count us.>That's what's so
frustrating!!!it is?>Dik Winter and Nora Baron are the two
most capable, most able at>little tricks like that, where they
hide their arguments in stuff that>you should catch, like that
x(x+1)/2 is always true in integers>because integers have
uniqueness of factorization.I thought uniqueness had
somethingto do with primes. this sounds like sour grapes.>But
algebraic integers do NOT!No ... but if A is an alg integer,is
it true or not that A(A+1)/2 is also?I dunno. Prob not. who
says one way orthe other?>The ideas of Dik Winter and Nora
Baron do NOT pass the smell-test.they smell OK to me. unliek
some others.>Ultimately they're relying on confusion, and that
you don't pay>careful attention so that they can sneak in
little tricks like>x(x+1)/2 in integers.how is it a trick?
isn't it true?>Without uniqueness of factorizaton Dik Winter
can't give a reason why>for any particular w_1(x) and w_2(x),
and notice he doesn't even>try!!!Dik gave explicit
definitions. How is unique factorization involved?and anyway,
what do you mean? a reason why what?>Why should he?why should
he what?>As long as I'm supposedly just some nut, you are some
nut. no doubt on that.and you're just willing to>attack my
ideas without wondering about the underlying mathematics,
as>long as a dumb device like x(x+1)/2 in the ring of integers
get's past>you, why should he worry?where did he use this
device?>Meanwhile he keeps up his own personal attacks with a
webpage harping>on an old abandoned argument of mine from
YEARS ago on x^4 + y^4 =>z^4.>The guy is SCUM. pond-scum?And
he's fooled you repeatedly, and that's why Iadmittedly he is
damn good at math and clearly youare not. what else is
new???Mr. Don't squeeze the charmin' Whipple>James
Harris===
===
Subject:
: How many long primes are there?
Mime-version: 1.0Content-type: text/plain;
charset=US-ASCIIContent-transferncoding: 7bitN can represent
any integerL represents the length of the period of the
expansion of 1/NThe residues at the same level have the same
period lengthsFor each subsequent residue the period length
halves until the length isodd. At that level the residues form
a cycle.Can anyone tell me how many long primes there are to
various bases?Can anyone tell me what was the deep work that
Christopher Hooley has donethat John Conway and Richard Guy
alludes to in The Book of NumbersWhy didn't they mention the
correlation of period length to the quadraticgraphs below?
They mentioned the Totient rule and Wilsons theorem. Is
thistoo basic?Could someone enlighten me pleaseN L Graph of
quadratic residues b where b also represents the base in which
the expansion of 1/n occurs2 1 1 3 2 2 1 1 4 2 3 1 1 5 4 |3|2|
2 4 1 1 7 6 |3|5| 3 | |2=4| 11 10 |7|6|8|2| 5 |5=3=9=4| 2 11 1
1 13 12 |2|11|6|7| 6 | 4 | 10| 3 | 3 = 9 | 4 |8|5| 2 |12 | 1 |
1 | 17 16 |6|11| 7|10|5|12|3|14| 8 | 2 | 15 | 8 | 9 | 4 | 4 |
13 | 2 | 16 | 1 | 1 | 19 18 |14|13|2|15|3|10| 9 | 6=17=4=16=9=
5| 6 |12|8| 3 |11=7| 2 |18| 1 |1| 23 22
|5|21|19|7|20|14|11|17|10|15| 11 |2= 4=16=3= 9=12= 6=13= 8=18|
2 |22| 1 | 1| 29 28 |2|27|10|19|11|18| 14 | 4 | 13 | 5 | 7 |16
= 24 =25 | 28 |3|26|8|21||14|15| 14 | 9 | 6 | 22 | 7 |23 = 7 =
20 | 4 |2|17| 2 | 28 | 1 | 1 | 31 30 |3|22|12|11| 15
|9=19=20=28| 30 |21|24|13|17| 15 |7=18=14= 10| 10
|23|29|27|15| 5 |2= 4=16= 8| 6 |6|26| 3 |5=25| 2 |30| 1 | 1|
===
===
Subject:
: Huge PerturbationsHiGiven a matrix A + b B, where
A and B are of thesame order, b > 1, and B has a null
spacedimensionality > 1,I'm interested in both the eigenvalues
of A + b B, androots of (A + b B) x = c (c is of the same order
as Aand B) : the asymptotic case b -> inf and how
it'sapproached.Can anyone point me to the relevant literature
or keywordsto search for? (I already tried the
obvious)===
===
Subject:
: Re: strain softening spring
<q6UzQPh=uis55kMn8JErWu9jUFJU@4ax.com>
<ec4c35fe.0402191025.ffc01e4@posting.google.com>i.e. the
equation I am trying to solve is as follows:>y'' + cy' +
[ko*e^(-alpha*t)]y = 0>Any help would be most
appreciated.>sincerely>Paul Joseph>(pjoseph@excite.com)Maple
gives a solution in terms of Bessel functions. I posted a
filecalled DE.pdf showing the solution
at:http://math.asu.edu/~kurtz/de/--Lynn> Hi Lynn,> This is in
regard to the solution you kindly posted above.> Can the first
argument of the BesselY function be negative?Now, having done
the required manipulations by hand[1], I getu^2 d^2f/du^2 + u
df/du + (b^2 - u^2) = 0,whereu = A(k,a,c)*exp(-at/2), b = c/a,
f(t) = y(t)*exp(ct/2)and A(k,a,c) is whatever it is, which has
solutions J_b(u) and Y_b(u).In any case, I'm not sure that
exp(-ct/2)Y(A(k,a,c)*exp(-at/2)) isactually bounded as t ->
+oo for all positive k,a, and c, so that thissolution is
therefore unphysical.> I tried to implement this in Excel, but
Excel's BesselY function appears> to require that the first
argument be positive. I am not sure if this> is a quirk of
Excel... or whether this is a mathematical
requirement.Conventionally[2], J_b(x) and Y_b(x) are indexed
with positive b.[1]: The original equation was a linear ODE in
one unknown, with onenon-constant coefficient. It should not be
necessary to throw this at acomputer to solve it, unless you
want the actual numerical values.[2]: In the UK, at least.--
P.A.C. SmithThe vast majority of Iraqis want to live in a
peaceful, free world.And we will find these people and we will
bring them to justice.===
===
Subject:
: Re: No Set Contains Every
Computable Natural> Given any set, S, of representations of
natural numbers,> I can write a TM that will find a
representation of a natural number> that is not in S. (No,
this TM does not halt after a finite number> of steps.)>This
is, of course, simply false, false, false.>If you give the set
of all representations of natural numbers, then>your TM will
not find any representation of a natural number not in>your
set.>You're speaking nonsense again.>Of course, we must be
careful here and fix a convention for when a>tape represents a
set of natural numbers. Let's say that a tape>represents a set
of natural numbers if it consists of blocks of ones>separated
by a single space.>We must also be careful to require that
your TM has the property that,>for each square x of the tape,
there exists a steps t such that for all>steps s after t, the
symbol on the square x at time s is the same as>the symbol on
x at t. After all, a TM which repeatedly changes the>0th
square to 1 and blank and 1 and blank cannot be said to
produce>any tape at all.I have given this proof several
times.> Yes, and you've been simply wrong several times.My TM
has the property you describe.>Now, on an input tape
containing 1, ,1,1, ,1,1,1, ,1,1,1,1, ,...,How about
b1b11b111b1111...>there is simply no possibility that your TM,
even after an infinite>number of steps, creates a tape with a
representation of some natural>number not on my input tape.
Every natural number is represented on>my tape. None is
missing. Your tape either returns a set of numbers>already on
my tape or it returns no set of numbers at all.This TM will
find a representation not on your tape:It is a three state
machine and I can provide a statetransition table if you
like.1) Find a blank2) Find a second blank3) Backup and write
a 1 on the previous blankrepeat steps (1) through (3)This TM
will produce a tape that contains exactly one blank.The
contiguous string of 1's preceding this blank will bea
representation not on your tape.> It will produce no such
tape. It will produce a tape consisting of> all 1's. This is
obvious.> Suppose that it contains a blank. Then that blank
must occur at some> square on the tape, say square n. It is
trivial to see that, by the> nth iteration of your three
steps, square n is no longer blank.> (After the first
iteration, square 1 is not blank, square two was> never blank,
square three is not blank after the second iteration and> hence
is also not blank after the third, and so on.)This TM always
checks to see if there is another blank on the tapebefore
overwriting the previous blank. That is what step 2 does.It is
easy to prove there is a blank on the output tape.> Your proof
of this claim is simply another confusion.It is a three state
TM.How confusing can it be?>No, no, no. A TM cannot extend the
representation of natural numbers>for two reasons: (1) It's a
ing Turing machine, isn't it? I don't>want to argue about
requirements for intentionality or whether TMs are>capable of
intelligence, but this TM is not part of the
negotiations>regarding our conventions.No intelligence
required.My TM just adds all the numberstogether (plus 1 for
each addition).A TM is too stupid to know thatthe sum is
supposed to be infinity.> If your tape contains a single
contiguous block of an infinite number> of 1's, then your tape
does not contain a set of natural numbers.A single contiguous
block of an infinite number of 1's followed bya blank in a
finite position? I never said the output of my TMcontained the
set of all natural numbers. I said it would containa
representation not on the initial input tape. The output
containsthe representation of exactly one natural number.>
Nothing can change that fact *except* changing what our
conventions of> translating tapes to subsets of N and back
are. Your machine is> invoked only after the conventions are
set and cannot change any> conventions.Namely, each natural is
represented by a contiguous string of 1'sfollowed by a blank.
This TM produces such a representation.> It is obvious that
your machine does not produce a (tape representing> a) set of
natural numbers even though at each finite step the tape the>
machine is working on represents a set of natural numbers.My
TM produces a tape with one representation of a natural
number.This representation is not on the initial input
tape.>(2) Every natural number has a>representation in our
convention, so the representation cannot be>non-redundantly
extended.I just described such a TM.No tape can contain every
representation of a natural number.There will always be a
bigger representation.> You have not extended the
representation of N. You haven't even> described correctly
what tape your machine produces. If you could> describe what
your machine produces and if you were capable of>
understanding, then you would realize that your machine does
not> produce a tape representing a set of natural numbers.This
TM produces a tape with a contiguous string of 1'sfollowed by a
blank. The output tape contains the representationof exactly
one natural number.Russell- 2 many 2 count===
===
Subject:
: Klingon,
UG, et UGG (Re: the anticlassicalist }{ i: linguistic
negation)> Presumably the child would re-structure it into>
something else which conformed with UG. Presumably :-) Into
Neanderthalish perhaps?They say there are lots of UGs in it.
Even UGGs.And GUGGs (Generalized Universal
GrammarGlossematics, or whatever fancy jargon strikesyours
[*])> However, from the examples> given, it appears as Jacques
says that the counter-universal features> are rather trivial or
perhaps not counter-universal at all.Nothing counter-universal
at all. Only one thing, mentionedin the grammatical part of
the dictionary... I don'tremember it clearly, and I just don't
feel likehunting for it now. If memory serves vaguely
enough,there is a certain type of relative clause which,either
cannot be expressed, or cannot be disambiguated.It is not there
by design, however, but by oversight.It has to do with two
syntactic rules clashing, so thatyou cannot properly hook your
relative clause onto the main clause. Big deal. Alas, the poor
Yorick whom I knew ...Alas, that poor Yorick (I knew him)...
will do just asnicely, thank you very much.[*] fancy, of
course.===
===
Subject:
: Re: Solving Equation with ln> Humm, I am
not familier with the Lambert W function, is there a web> site
you would recommend so that I can read up on it or is this>
function too complex for the non-math major type person.I'll
let you decide whether it's too complex for your taste.
Briefly,it's just the inverse of the function f(W) = W*e^W.
(Although that inverseis multivalued, you only need to use the
principal branch for solving yourequation.) See
<http://mathworld.wolfram.com/LambertW-Function.html> formore
information, including series expansions, etc.David> I was
working on a math problem at work and ran into something I>
don't know how to solve and was hoping someone could give me a
clue> on this fairly simple problem.> Original Equation:>
227.11 = 5.8{ln(38/x)} + (38-x)> Simplfied Equation:> 189.11 =
5.8*{ln(38/x)} - x> of course you can reduce this down to>
32.60 = ln(38/x) - x/5.8> and further (I think) to> 1.44 x
10^14 = 3.8/x - e^(x/5.8)> but I am stuck after this. Is it
possible to solve for X?> Not in terms of the standard
elementary functions. But youcan solve in terms of the Lambert
W function...x = W(38*exp(-(189.11)/5.8)/5.8)*5.8 =
2.627466197*10^(-13);===
===
Subject:
: Re: Are the derivatives of
abs[(x-a)^3] different for x>a and x<a ?
<Pine.SOL.4.44.0402191329400.8274-100000@red.csi.cam.ac.uk>