HDTenEightyP/NetNews huggingface dataset:
From: naturman@uctvax.uct.ac.za
Newsgroups: sci.math
Subject: Mathematical foundations of QM
Keywords: QM
Date: 28 Dec 92 17:45:16 GMT
Article-I.D.: uctvax.1992Dec28.154516.203431
Organization: University of Cape Town

The following diagram shows how various mathematical formalisms for QM are
related. 

Wave Mechanics            Matrix Mechanics
     ^                          ^
     |                          |
     |                          |
     ----------------------------------------------->Fringe approaches:
                  |                                   Convex sets
                  |                                   Transition probabilities
           Hilbert space QM                           Involution algebras
                  ^  ^
                  |   \
                  |    \
                  |   Orthomodular lattice approach   
                  |     ^
                  |    /
                  |   /
        Geometry/closure space approach


Where approaches at the bottom of an arrow can be regarded as more fundamental
than approaches at the end of an arrow.

I am currently working on an approach using orthogonality relations on sets of
states. So far it looks like it could turn out to be more fundamental than the
closure space approach. I think that Chaos theory will be required to justify
the concept of "quantum states" which is taken as a primative concept in most
of the above approaches.
From: pete@bignode.equinox.gen.nz (Pete Moore)
Subject: Re: Multiples of 9
Newsgroups: sci.math
Organization: Equinox Networks
Reply-To: pete@bignode.equinox.gen.nz
X-Newsreader: TIN [version 1.1 PL8]
Date: 21 Dec 92 16:51:22 +1200
Organization: Equinox Networks

mingos@vax.oxford.ac.uk wrote:

>  I hope I'm posting this into the correct maths group, but I have a problem
>which has started bugging me, it may be an old chestnut, but I can't find a 
>solution that satisfies me.

>  The basic thing is that if you multiply any number by 9, you end up with a
>number whose digits add up to 9, OR whose digits add up to a number whose
>digits add up to 9, OR...etc, depending on how big the number is.
>  so, for example, 9x12343 = 111087.  1+1+1+0+8+7 = 18.  1+8 = 9.

>  I can figure a real hand waving argument for this, but does anyone have a
>more definitive proof, or can point me towards one?  I'm a chemist by
>background, so I don't know my way around the mathematical lit.  I don't even
>know which branch of mathematics this is...

It's number theory.

First I'll explain congruences, in case you aren't familiar with them.
I'll use # for the congruence symbol (nomally written like = but with 3
horizontal lines instead of 2), i.e. I'll write
  a # b (mod c)
to mean that a - b is an integer multiple of c. In particular, a#0 (mod 9)
if and only if a is a multiple of 9.

Some relevant properties of congruences:
  (1) a + b  #  {a (mod c)} + {b (mod c)}   (mod c)
  (2) a * b  #  {a (mod c)} * {b (mod c)}   (mod c)
  (3) 10 # 1 (mod 9)  and so  10^k # 1 (mod 9) for any positive
             integer k, by applying (2) repeatedly

Suppose n has k base-10 digits. Let these digits (in order) be  a_1, a_2,
 ... a_k

Then  n = a_1*10^{k-1} + a_2*10^{k-2} + ... + a_{k-1}*10 +a_k

Using (1)-(4) above, n # a_1 + a_2 + ... + a_k  (mod 9)  and hence n is
divisible by 9 (i.e. n#0 (mod 9)) if and only if the sum of the digits of n
is divisible by 9 (and this sum is divisible by 9 iff the sum of _its_
digits is divisible by 9, etc).

Example:
 
  let   n = 5283 then a_1=5, a_2=2, a_3=8, a_4=3,
   and  n = 5*10^3 + 2*10^2 + 8*10 + 3
   so   n # 5+2+8+3 = 18 (mod 9), since the powers of 10 are # 1 (mod 9).
   But  18#0 (mod 9), and so 5283 _is_ divisible by 9.
--
 +-------------------  pete@bignode.equinox.gen.nz  -------------------+
 | The effort to understand the universe is one of the very few things |
 | that lifts human life above the level of farce, and gives it some   |  
 | of the grace of tragedy   -  Steven Weinberg                        |
 +---------------------------------------------------------------------+
Newsgroups: sci.math
From: asimov@wk223.nas.nasa.gov (Daniel A. Asimov)
Subject: Re: The Continuum Hypothesis:  Must it be {True or False}, or Not?
Organization: NAS, NASA Ames Research Center, Moffett Field, California
Date: Thu, 24 Dec 92 17:59:30 GMT

In article <1992Dec24.034938.11339@smsc.sony.com> markc@smsc.sony.com (Mark Corscadden) writes:
>In article <1992Dec14.200024.6435@nas.nasa.gov> asimov@wk223.nas.nasa.gov (Daniel A. Asimov) writes:
>>       Do people buy this, that FLT must be {true or false}, regardless of
>>       whether it is provably true or false (i.e., decidable)?
>>Dan Asimov
>
>In email, Dan suggested using the phrase "having determinate truth-value"
>to say that a proposition must be {true or false}, regardless of whether
>it is provably true or false.
>
>I'll shorten this, simply saying that such a proposition is "determinate".
>
>To answer Dan's question, I buy - beyond any doubt - that FLT is
>determinate.  On the other hand I can't convince myself that CH (the
>Continuum Hypothesis) is determinate, but at the same time I can't
>find any necessary reason to believe that CH is indeterminate either.
>
> [...]
>Mark Corscadden
-----------------------------------------------------------------------------------

I would like to comment on what I perceive to be the difference between 
the likelihood of FLT being determinate versus the likelihood of CH being 
determinate.

Time.

That is, how much "time" would it take to check either of these propositions?

In the case of FLT, we could completely check its truthfulness in countable time:
enumerate all candidates a,b,c,n, and do the arithmetic. After countable time is over,
we know the answer. (Here the "unit" of time is the time it takes to calculate
a^n + b^n - c^n and decide whether it is equal to 0.)

In the case of CH, there are a lot more things to check:  We can imagine considering
all possible infinite subsets S of the reals R--all 2^c of them.  For each such S, we 
can imagine all possible mappings f: S -> R, and all possible mappings g: S -> Z from
S to the integers.  And we can imagine checking whether f or g is bijective.
If, for each S, either an f: S -> R or a g: S -> Z is bijective, then of course 
CH is true; otherwise it's false.

An unthinkable number of things to check.  However, the whole process could be broken 
down into tiny, easily-imaginable steps:  LOTS of them.  The set of all steps to carry
out could be well-ordered.  Then the steps could be (tediously) carried out in a very
large cardinal (or shall we say ordinal?) amount of time.

So IMHO, the salient difference between the likelihood of FLT's being determinate,
and CH's being determinate, is that CH would simply take a *lot longer* to check.
But since I can clearly conceive of carrying out either check, I personally believe
that both of them are determinate.  

(Although in the case of CH, given the chance I would definitely delegate the 
responsibility of carrying out the check. (:-)>)

Dan Asimov
Mail Stop T045-1
NASA Ames Research Center
Moffett Field, CA 94035-1000

asimov@nas.nasa.gov
(415) 604-4799
Newsgroups: sci.math
From: edstrom@Elmer.hsc.ucalgary.ca (John Edstrom)
Subject: references request
Date: Mon, 28 Dec 92 17:33:49 GMT
Organization: Neuroscience Division, U of Calgary School of Medicine, Calgary, Alberta, Canada

Does anyone know of a general, introductory level review or discussion
on strictly sofic systems and subshifts of finite type?  I'm
particularly interested in how, when and where they are used.

Also, what is "ultrametricity?"

John Edstrom | edstrom @ elmer.hsc.ucalgary.ca



--
 RM 2104, HSc Building,  Div. Neuroscience
 U. Calgary School of Medicine,  3330 Hospital Drive NW
 Calgary, Alberta       T2N 3Y4
 (403) 220 4493  (wk)
Newsgroups: sci.math
From: yoparish@iiic.ethz.ch (Yoav Ilan Haim Parish)
Subject: Two (I think) interesting Problems
Originator: yoparish@a12
Organization: Dept. Informatik, Swiss Federal Institute of Technology (ETH)
Date: Wed, 23 Dec 1992 13:46:51 GMT

I have (had) the following problems which I`d like to pose in public:

	1.) How can I fold a pentagon from a quadratic sheet of paper easily ?
	    How can I fold an approximate (regular) pentagon from a quadratic sheet of
		paper easily ?


	2.) 	On a sheet with quadrats on it (as used in schools for writing)
		I'd like to draw an equal sided triangle where every one of the	
		three points should be on a edge of a quadrat.
		If this is not possible, why not ?
		How can I draw a good approximation of an equal sided triangle
		with every on of the three points on the edge of a quadrat.
		(Make use of the fact that the human eye doesn't see horizontal
		and vertical lines at same sizes)


I dont know the answers. If you have any or find any I would be glad if you could
send them to me by e-mail (if nobody else is interested)

Thanks
yogi

92/93

Newsgroups: sci.math
From: dy@shire.math.columbia.edu (Deane Yang)
Subject: Biographical sketch of Mikhael Gromov
Organization: Mathematics Department, Columbia University
Date: Tue, 22 Dec 1992 20:01:45 GMT
X-Posted-From: shire.math.columbia.edu
NNTP-Posting-Host: sol.ctr.columbia.edu


I need ASAP a biographical sketch of Mikhael Gromov.
Does anyone know where I can find one? For example,
if he's won a prize, then the announcement of the prize
would contain a sketch. Or can someone provide a short
history of his career? I know a rough outline, but would
appreciate more information.

Deane Yang
Polytechnic University
Xref: sparky sci.math:17436 alt.books.technical:395
From: erik@naggum.no (Erik Naggum)
Newsgroups: sci.math,alt.books.technical
Subject: Re: High Prices of Math Books. I am pissed.
Date: 27 Dec 92 16:08:57 GMT
Reply-To: Erik Naggum <enag@ifi.uio.no>

Graeme Hirst <gh@cs.toronto.edu> writes:
:
|   In particular, they will sell a cheap paperback edition only outside
|   North America, while insisting that North American customers buy the
|   hardcover edition at a much higher price.

Tell you what, those hardcover books that you think are so expensive
cost us Europeans twice as much.  _That's_ why we get those "cheap"
paperbacks which only cost about as much as your hardcopy, i.e. _half_
the price -- a bargain!

Best regards,
</Erik>
--
Erik Naggum                 ISO  8879 SGML                    +47 295 0313
                            ISO 10744 HyTime
<erik@naggum.no>            ISO  9899 C                 Memento, terrigena
<SGML@ifi.uio.no>           ISO 10646 UCS             Memento, vita brevis
Newsgroups: sci.math
Subject: Re: TO ALL !!! -- End of Discussion, I hope
From: a_rubin@dsg4.dse.beckman.com (Arthur Rubin)
Date: 21 Dec 92 23:07:43 GMT
Organization: Beckman Instruments, Inc.

In <1h323mINN1rt@usenet.INS.CWRU.Edu> ak005@cleveland.Freenet.Edu (Yona Webb) writes:


>My posting (Dec. 18) has offended many netters.  Several people
>expressed quite vocally their anger and displeasure, both publicly and in
>E-Mail. There was no need for flames however, just to let me know that
>I have made a mistake.
> 
>Why did I send the card in public posting to Sci.Math?  -- About a month
>ago I posted a msg in Sci.Math seaking assistance on couple of topics.
>A dozen of people replied!  They offered help, useful info and assured
>me that they would be willing to help in the future, if the need arises.
>A private  -Thank You Msg-  was sent to all of them.
> 
>There has been a lot of excitement in the air in the last week or so.
>I made  *ONE*  card to reflect this excitement and to greet all my friends,
>Christian and Jewish.  At one moment I thought of Sci.Math. as a single
>entity because of the help I received from so many people there and ... hop,
>the card was uploaded in a public msg TO ALL!!! Good intentions, wrong move!
> 

May I suggest, in this newsgroup, a graphic of an apple (not Apple (TM)), with
an announcement:  Happy Newton Day.

(Some readers may recall that Sir Issac Newton was born on December 25.)

--
Arthur L. Rubin: a_rubin@dsg4.dse.beckman.com (work) Beckman Instruments/Brea
216-5888@mcimail.com 70707.453@compuserve.com arthur@pnet01.cts.com (personal)
My opinions are my own, and do not represent those of my employer.
My interaction with our news system is unstable; please mail anything important.
Xref: sparky sci.math:17327 rec.puzzles:8101
From: jbaez@riesz.mit.edu (John C. Baez)
Newsgroups: sci.math,rec.puzzles
Subject: Re: Marilyn Vos Savant's error?
Keywords: savant
Date: 22 Dec 92 00:36:23 GMT
Organization: MIT Department of Mathematics, Cambridge, MA

In article <2B30BAC8.26585@news.service.uci.edu> srw@horus.ps.uci.edu (Steven White) writes:
>In article <1992Dec17.041022.29031@thunder.mcrcim.mcgill.edu>, mouse@thunder.mcrcim.mcgill.edu (der Mouse) writes:
>|> In article <1992Dec15.055832.26324@galois.mit.edu>, jbaez@riesz.mit.edu (John C. Baez) writes:
>|> 
>|> > 1)  You draw 4 cards from a well-shuffled standard deck.  You turn
>|> >     one over and it's an ace.  What's the probability that they are
>|> >     all aces?
>|> 
>|> There are 4! C(52,4) = 52!/48! = 6497400 sets of 4 cards you could
>|> draw, counting different orders as different.  In one case in 13 you'll
>|> turn up an ace, for a total of 499800, while in only 24 cases are they
>|> all aces.  The answer to this question is therefore 24/499800, which is
>|> 1 in 20825, or, to 20 places from dc, .00004801920768307322.
>|> 
>|> > 2)   You draw 4 cards from a well-shuffled standard deck.  You turn
>|> >      one over and it's the ace of hearts.  What's the probability
>|> >      that they are all aces?
>|> 
>|> Of the 6497400 sets of 4 cards, in one case in 52 will you see the ace
>|> of hearts, for a total of 124950.  Again, 24 are all aces.  The answer
>|> is thus 24/124950, or 12/62475, about .00019207683073229291.  This is
>|> roughly four times as large as the previous answer.
>|> 
>|> 					der Mouse
>
>No way for 2). 24 possibilites are all aces, but only 6 of those had the
>ace of hearts as the first card. Thus 6/124950=1/20825, same as 1.
>Its clear if you think about the questions a little more that 1 and
>2 have to have the same answer.

Steve's right.  Let me do, out loud, the thinking Steve suggests.
Suppose you draw four cards, turn one over, and its an ace.  There are
four cases - hearts, diamonds, clubs, spades.  Let p_i be the
probability that one has drawn 4 aces, *given* that the ith case has
occured (i = hearts, diamonds, clubs, spades).  By symmetry all p_i are
equal, say p.  Now turn to problem 1: one has drawn four cards, turned
one over, and its an ace.  The chance that the ith case has occured is
1/4, by symmetry.  So the answer to problem 1 is p/4 + p/4 + p/4 + p/4 =
p.

Of course, its reasoning along these lines, but ERRONEOUS reasoning,
that might make one suspect the answers to my "improved" pair of
questions were also equal.
Newsgroups: sci.math
From: hrubin@pop.stat.purdue.edu (Herman Rubin)
Subject: Re: Lagrange successor rule
Organization: Purdue University Statistics Department
Date: Wed, 23 Dec 1992 16:04:37 GMT

In article <3207@devnull.mpd.tandem.com> garyb@anasazi.UUCP (Gary Bjerke) writes:
>
>I was thumbing through an old statistics textbook when I came across the 
>Lagrange successor rule. The example given was that of a collection of coins
>for which the probability of flipping a head is uniformly distributed over
>the set of values {1/N, 2/N, ..., N/N} for N some arbitrary integer. The rule
>states that the probability of getting a head on the (n+1)th flip given that
>the first n flips were heads, is n/(n+1).

You left out 0/N, but this can be done in many ways.  

>Note that N is arbitrary and not a parameter is the final result. The textbook
>said Lagrange had used this to prove that the sun was extremely likely to rise 
>tomorrow, given that it had risen every day for the past 5000 years (the 
>probability is [5000*365]/[5000*365+1], to be precise). 

>I followed the proof, but I have absolutely no intuition for this result at
>all. I even fail to see how it applies to the rising of the sun (in what sense
>does the unconditional probability of its rising meet the uniform-distribution
>requirements?) Can somebody help me get a gut feel for what this result means?

I thought it was Laplace who said this, but anyhow this is a somewhat standard
misuse of Bayes' Theorem.  For the argument as usually given to be valid, the
prior probability that the sun would rise on a random day would have to be 
uniformly distributed between 0 and 1.  This version of the "principle of
insufficent reason" is no more appropriate than any of the other versions,
and is a common source of paradoxes and fallacies.

-- 
Herman Rubin, Dept. of Statistics, Purdue Univ., West Lafayette IN47907-1399
Phone: (317)494-6054
hrubin@snap.stat.purdue.edu (Internet, bitnet)  
{purdue,pur-ee}!snap.stat!hrubin(UUCP)
Xref: sparky sci.math:17414 sci.physics:21760
From: paul@mtnmath.UUCP (Paul Budnik)
Newsgroups: sci.math,sci.physics
Subject: Re: Bayes' theorem and QM
Date: 25 Dec 92 16:57:41 GMT
Followup-To: sci.math
Organization: Mountain Math Software, P. O. Box 2124, Saratoga. CA 95070

In article <1992Dec24.101452.16194@oracorp.com>, daryl@oracorp.com (Daryl McCullough) writes:
[...]
> Nonclassical measure theory simply drops the assumption that the
> measurable sets form a sigma algebra. It allows for the possibility
> that proposition A may have a well-defined probability, and proposition
> B may also have a well-defined probability, while the proposition A and
> B may be given no probability whatsoever.
> 
> The applicability to quantum mechanics should be obvious: wave
> functions allow us to calculate the probability that the particle can
> be found in region A, and they allow us to calculate the probability
> that its momentum is in region B, but they don't allow us to calculate
> the probability that the particle is in region A *and* its momentum is
> in region B. Quantum mechanics does not give a unique answer to such
> questions. [...]

The same thing is true of the frequency and location of a wave in classical
mechanics for a similar reason. Only a delta function has a 
precise location and its spectrum is flat over all frequencies.
We see no reason to change probability theory to accommodate that case
and I do not think there is any reason for doing so in QM.
Of course in QM we force the wave to be more or less like a delta function
by the type of experiment we perform on it. It would be
mistake to think that this operation of the experimental apparatus on
physical phenomena requires a new version of probability theory.

Paul Budnik
Newsgroups: sci.math
From: minsky@media.mit.edu (Marvin Minsky)
Subject: Re: Two (I think) interesting Problems
Organization: MIT Media Laboratory
Date: Mon, 28 Dec 1992 03:14:41 GMT

In article <1992Dec28.013845.23064@infodev.cam.ac.uk> gjm11@cus.cam.ac.uk (G.J. McCaughan) writes:
>1. To fold a sheet of paper into a regular pentagon: If it's an A4 sheet
>of paper (or any other of the A... sizes; what matters is that the ratio
>of the sides should be root(2)) there is a fairly elegant way to do this
>with perfect accuracy. It goes like this:

Fascinating.  We unhappy americans have no convenient source of
root(2) paper.  I suppose we could ask you to fax us a sample.  Or we
could cut out the heart of one of those xmas stars that are popular
around this time of year.  

ANother way to make a "perfect" regular pentagon is to cut a
rectangular strip of paper and simply tie it into a simple overhand
knot.  Then pull-fold-pull-etc. it until it is flat and tight.  Some
people find this surprising, having never observed this 5-sided
character of a knotted strip.

.
Newsgroups: sci.math
From: leife@ii.uib.no (Leif Erik R. Hauge)
Newsgroups: sci.math
Subject: Gaussian sums
Date: 30 Dec 92 13:04:10 GMT
Organization: Institutt for Informatikk, UiB, Bergen, Norge


	---  GAUSSIAN SUMS ---

I need to know something about the Gaussian sums.
What are they?  How are they defined?  To which
part of math do they belong (algebra? Numb. theory?)

I would also like to read an easy introduction to the
theory of Gaussian sums.  Do you know some standard textbooks 
which give such an introduction?

Any help appreciated!

	Erik
Newsgroups: sci.math,sci.physics
From: markc@smsc.sony.com (Mark Corscadden)
Subject: Re: Bayes' theorem and QM
Organization: Sony Microsystems Corp, San Jose, CA
Date: Thu, 31 Dec 92 02:40:39 GMT

In article <1ht1arINNf8a@chnews.intel.com> bhoughto@sedona.intel.com (Blair P. Houghton) writes:
>The two spheres thus formed do indeed have the same volume
>sum as the original sphere, but each has less volume than
>the original, or else they weren't composed of a finite
>number of pieces[*], or else when constructed they contained
>gaps, internally.

Blair, there's an excellent discussion of this in the sci.math faq:
17Q.- Cutting a sphere into pieces of larger volume  

Mark Corscadden
markc@smsc.sony.com
work: (408)944-4086
Newsgroups: sci.math
From: hougen@focus.csl.uiuc.edu (Darrell Roy Hougen)
Newsgroups: sci.math
Subject: Generalization of binomial theorem wanted
Date: 4 Jan 1993 05:26:46 GMT
Organization: Center for Reliable and High-Performance Computing, University of Illinois at Urbana-Champaign
NNTP-Posting-Host: focus.csl.uiuc.edu
Summary: Generalization of binomial theorem wanted
Keywords: binomial

In terms of m and n, how many unique terms are there in
	(x_1 + x_2 + ... + x_m)^n
when it is expanded?

I have a rather cumbersome formula and I'm looking for something more
concise.  This must be a FAQ but it wasn't in the sci.math FAQ file.

The formula I have is

                n 
            sum                 sum                 1  
            i = 1    k_1 <= k_2 <= ... <= k_i <= m

Darrell R. Hougen
Newsgroups: sci.math,sci.physics
From: freier@mail.physics.utah.edu (rodney james freier)
Subject: Re: What waves in light waves? (was Re: TIME HAS INERTIA)
Organization: University of Utah - Physics Department
Date: Sat, 26 Dec 92 07:54:45 GMT

In article <DREIER.92Dec18100700@durban.berkeley.edu>, dreier@durban.berkeley.edu (Roland Dreier) writes:
|> In article <Bz81F4.Lqt@quake.sylmar.ca.us> brian@quake.sylmar.ca.us (Brian K. Yoder) writes:
|> 
|>    A similar problem exists for physicists who consider light to be a wave (or
|>    a wave and something else too).  Waving is something an entity does, not a
|>    thing in itself.  There can no more be a wave traveling through space than
|>    there can be a room with some "jumping" going on outside of any entities
|>    which are doing the jumping.
|> 
|> Even more troubling for me, as a student of modern algebraic geometry,
|> is the concept of schemes.  How can one have a "scheme" in the absence
|> of someone scheming?
|> --
|> Roland Dreier                                        dreier@math.berkeley.edu

   Ahem.   There IS an entity 'waving'.  That entity is called the electromagnetic
field.  The whole point of identifying light as a wave is showing that there
is a "thing" (albeit a strange thing, especially when you consider quantization)
called the EM field that can store energy, propogate information, and... well,
act like a 'thing'.  

   The problem lies with considering the vacuum to be nothing.  The fabric of
space-time is NOT nothing.

Rod
(freier@mail.physics.utah.edu)