HDTenEightyP/NetNews huggingface dataset:
From: dream!Marcus_Brodeur@bikini.cis.ufl.edu (Marcus Brodeur)
Subject: PUBL: Amiga VR Mailing List
Originator: hlab@stein.u.washington.edu
Organization: DreamStates BBS & CyberNet HQ in G'ville, FL   1-904-331-4317
Date: Tue, 15 Dec 1992 15:22:27 GMT



 U>         I was just wondering, does anyone know if any software is
 U> available via FTP or otherwise for Amiga VR applications, such as a
 U> PowerGlove controller and or software that makes use of it?  How about
 U> the Sega glasses?  Anything else I may have no idea about?

Well, I suppose now is as good a time as any to mention that (thanks to the
suggestion of Richard A. Romano) a friend and I are setting up a mailing
list for people doing VR with Amigas.

Since my original post asking for a roll-call of Amiga VR people present, I
have gotten an incredible amount of response, and Richard's suggestion
seemed the most reasonable way of allowing Amiga folks the chance to
communicate without driving the rest of sci.v-w stark raving mad.

If you are interested, please send me some private e-mail at any of the
addresses below, and I'll add your name to the list.  Anyone *else* who is
interested in Amiga VR is welcome to drop me a line as well and be included
in the mailer.  Allow me to note that this is completely non-commercial in
nature; it's main purpose is to further communication between Amiga
implementors of VR to save everyone headaches...

 U>         I'm only running an A500 with 1 meg or RAM, so obviously the
 U> applications would be somewhat limited if possible at all.  I was
 U> thinking of getting a PC (normally against my religion) just for VR
 U> applications, but cash is somewhat of a problem at his point.
 U> However, if what I need is available for Amiga, I'd much rather just
 U> upgrade to what I'll need...

The applications are certainly possible.  Of course, it's a question of
*which* applications you'd want to be implementing in VR.  An increase in
memory is going to be called for.  But I wouldn't get a PC just for VR. 
The Amiga's sound and graphics capabilities are well documented, and
although there is not as much support at present as exists for PC's, there
*is* software out there (such as the Mandala program) for VR on the Amiga,
as well as many people working on it even as we speak (get on the mailing
list).

The decision is yours.  But you'll be able to make a more informed decision
if you talk to some of the many other Amiga VR people on the mailing list
once it gets going...

Hope to hear from you soon, as well as anyone else who is interested in
participating in the mailing list.

!!!!! T  E   M    P     U      S      F     U    G   I  T !!!!!

-- Via DLG Pro v0.995

* IF replying by MAIL, please reply to one of the addresses listed below...

  Internet: Marcus_Brodeur%dream.uucp@ufl.edu
        or: dream!Marcus_Brodeur@ufl.edu
      UUCP: uflorida!dream!Marcus_Brodeur
----------------------------------------------------------------------------

"Whistler?"  "I just want...peace on Earth and goodwill towards Man."
"Look, we're the government...we don't DO that sort of thing."  (Sneakers)
Newsgroups: sci.math
Subject: Re: Properties of Illumination Functions?
From: a_rubin@dsg4.dse.beckman.com (Arthur Rubin)
Date: 17 Dec 92 18:42:58 GMT
Organization: Beckman Instruments, Inc.

In <1992Dec16.225253.25173@jarvis.csri.toronto.edu> dret@dgp.toronto.edu (George Drettakis) writes:

>I sent this to sci.math.research, but I have a suspicion the big
>black net hole ate it. I apologize if you see this twice.
>-----------------------

>I have the following problem in computer graphics. Assume a rectangular 
>area in the plane z=d, with -xa1 < x1 < xa1 and -ya1 < y1 < ya1.
>This area is assumed to be a diffuse rectangular light source,
>with the surface normal 0, 0, -1.
>Assume also the receiving plane z=0, and the function I(x,y) defined on 
>this plane. The function I(x,y) expresses the illumination at this point:

>                       ya1    xa1
>                        /      /
>                       |      |                 1
>        I(x, y) :=     |      |   ----------------------------- dx1 dy1
>                       |      |            2           2    2 2
>                      /      /    ((x - x1)  + (y - y1)  + d )
>                    - ya1  - xa1

I was wrong.  It is obvious, by formal diffentiation, that I(X,Y) radially decreases.
--
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.
From: mingos@vax.oxford.ac.uk
Newsgroups: sci.math
Subject: Multiples of 9
Date: 18 Dec 92 16:52:48 GMT
Organization: Oxford University VAX 6620

Dear all,


  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...


					Thanks for any advice,
                                        (even if it's try a different group!)


							Gavin


******************************************************************************
Gavin Whittaker
Inorganic Chemistry Laboratory
South Parks Road
Oxford
England
*******************************************************************************
Xref: sparky sci.math:17069 rec.puzzles:7985 alt.usage.english:9624
From: bhoughto@sedona.intel.com (Blair P. Houghton)
Newsgroups: sci.math,rec.puzzles,alt.usage.english
Subject: Re: Naming Large Numbers (Re: Negative Zero)
Date: 17 Dec 1992 01:26:05 GMT
Organization: Intel Corp., Chandler, Arizona
NNTP-Posting-Host: stealth.intel.com

10^(10^126) pi

The other answer is:  you've given two "large" numbers that
are also unknown in value, so of course one can't express
them in scientific notation except as convenient symbols.

				--Blair
				  "680V790M;720V800Q800A"
Newsgroups: sci.math
From: palais@binah.cc.brandeis.edu
Subject: Re: Marilyn Vos Savant's error?
Reply-To: palais@binah.cc.brandeis.edu
Organization: Brandeis University
Date: Tue, 15 Dec 1992 01:24:03 GMT


You have a hat in which there are three pancakes.  One is golden on
both sides, one is brown on both sides, and one is golden on one side
and brown on the other.  You withdraw one pancake and see that one side
is brown.  What is the probability that the other side is brown?

---Robert H. Batts, Acton, Mass.

It's two out of three.  The pancake you withdrew had to be one of only
two of them: the brown/golden one or the brown/brown one.  And of the
three brown sides you could be seeing, two of them also have brown on
the other side.
--
Dave Burchell                   |
burchell@cse.unl.edu            |        Review your options.
djburche@jwendnelnc.cr.usgs.gov |              Amiga.
=======
I think she is correct (as usual). It is clear that the pulling out of 
the hat involved choosing one of SIX sides to look at---all equally likely. 
Their are THREE brown sides (there are TWO ways you could have picked 
the brown/brown and ONE way the brown/golden---all three equally 
probable.)Of these three equally likely possibilities two have a 
brown other side. If you don't believe it, test it experiment 
experimentally a few dozen times and see if you come close to 66% 
or 50%.
    R. Palais
From: noffke@carroll1.cc.edu (Pat Noffke)
Newsgroups: sci.math
Subject: Interesting Discrete Math Question
Date: 15 Dec 92 21:19:57 GMT
Reply-To: noffke@carroll1.cc.edu (Pat Noffke)
Organization: Carroll College-Waukesha, WI

Here's a question we had on one of our last tests in Discrete.

A jar contains 100 chips numbered 1 through 100.  I want to draw a
subset of chips that satisfies the condition that at least two chips
have values that differ by exactly 3.  What is the minimum number of
chips I must draw to guarantee that I satisfy the condition?  Justify
your response.



That's it.  I already have one approach.  I'm just curious as to what
some others may be.  Thanks a lot.


Patrick Noffke
Newsgroups: sci.math
From: pfleury@ddsw1.mcs.com (Patrick J. Fleury)
Subject: Re:  Haughty quote
Date: Tue, 15 Dec 1992 01:28:21 GMT
Distribution: usa
Organization: ddsw1.MCS.COM Contributor, Chicago, IL

In article <1gb49eINN3ra@mathj.usc.edu> bruck@mathj.usc.edu (Ronald Bruck) writes:
>
>I like Spivak's sense of humor (esp. the "Institute of Haughty Attitudes"
>in his Joy of TeX), but let's get a little closer to source.  I'll quote from
>that notedly accurate (8-) historian of mathematics, Eric Temple Bell (Men
>of Mathematics, p. 452):
>
>	In passing it may be amusing to recall that Liouville inspired
>	William Thompson, Lord Kelvin, the famous Scottish physicist,
>	to one of the most satisfying definitions of a mathematician
>	that has ever been given.  "Do you know what a mathematician is?"
>	Kelvin once asked a class.  He stepped to the board and wrote
>	$$\int_{-\infty}^{+\infty} e^{-x^2}\,dx = \sqrt{\pi}.$$
>	Putting his finger on what he had written, he turned to the
>	class.  "A mathematician is one to whom {\it that\/} is as
>	obvious as that twice two makes four is to you.  Liouville was
>	a mathematician."
>
>But I believe Bell lifted this (almost verbatim!) from "Life of Lord
>Kelvin", by S. P. Thompson.  Robert Edouard Moritz, in his "On Mathematics
>and Mathematicians" (Dover) quotes a very similar passage in "Life of
>Lord Kelvin" (London, 1910, p. 1139).  (Perhaps biographers were more prolix
>in those days.)
>
>I very much doubt that Euler made a similar statement.  This is probably a
>case of faulty memory on the part of the high-school teacher.
>
>--Ron Bruck
>bruck@mtha.usc.edu preferred

Actually, I like what happened in Spivak a little more.  The quote appears 
at the end of the problem set of chapter 3 and in it _mathematician_ is
spelled wrong -- twice.

Pat F.
pfleury@ddsw1.mcs.com

Xref: sparky sci.math:17139 sci.physics:21412 sci.bio:4625 sci.chem:5493 misc.education:5403
From: hcart1@vax.oxford.ac.uk
Newsgroups: sci.math,sci.physics,sci.bio,sci.chem,misc.education
Subject: Re: TIME HAS INERTIA - BASS JEALOUS AND INSECURE-ABIAN VALIDATED
Date: 17 Dec 92 08:54:11 GMT
Organization: Oxford University VAX 6620

In article <abian.724200923@pv343f.vincent.iastate.edu>, abian@iastate.edu (Alexander Abian) writes:

> 
> C.R. Bass is discredited as a
> jealous, putting down envious person who gets his feeling of security
> by most pathetic means of eminating arrows of hate, sarcasm, enmity,
> hostility and antipathy.  
>     

This is not a helpful addition to the debate from Dr Abian....

>   
> in France and Germany there are conferences studying TIME HAS INERTIA
> topic and that THEY ARE MORE AND MORE CONVINCED THAT in all those
> cases where there is a PUSZZLING LOSS OF ENERGY -  ABIAN's (A2)  IS
> THE NSWER:

...but this is. Now at last we can have some concrete facts. Where are
these conferences, and who are these scientists who are now convinced
of the validity of these theories? Answers, please Dr Abian; many of us
would be only too delighted to have the chance to speak personally
to colleagues who understand your theory in detail.

Dr Hugh Cartwright
Physical Chemistry Laboratory
Oxford University, England.
Newsgroups: sci.math,rec.puzzles
From: snewman@Xenon.Stanford.EDU (Steven Newman)
Subject: Re: Marilyn Vos Savant's error?
Organization: Computer Science Department, Stanford University.
Date: Tue, 15 Dec 1992 20:02:56 GMT

> |>   1) You meet a man on the street and ask him how many children he has.
> |>      He replies "two, and one is a boy."  What is the probability that
> |>      his other child is also a boy?
> |>
> |> ...
> |>
> |>The answer to problem 1 is 1/3, while the answer to problem 2 is 1/2.
> |>(One must be careful in the interpretation of the statement in problem
> |>1; it means "I have two children, and it is not the case that both of
> |>them are girls.")
> 
> Sorry!  You made the same mistake as the first poster!
> The odds of the other child being a boy is 1/2 in both cases.
> Observe:
> 
> Case 1)  Child = boy   Other = ?
> Yes  ->  a) Child = boy , Other = boy
> No   ->  b) Child = boy , Other = girl
>          c) Child = girl, Other = boy
>          d) Child = girl, Other = girl
>          Obviously, c and d are invalid.
> 
> ...

This assumes that there is an explicit identification of one child as
being "the child", and the other as being "the other child", *before*
testing whether "the child" is a boy.  If you make this assumption, then
indeed the odds of "the other child" being a boy are 1/2.  However, the
assumption is invalid.  In case (c), the man could still give the
answer "two, and one is a boy"; here "one" would be "the other child",
not "the child".  So, under this analysis, case (c) is not invalid.

 - Steve Newman (snewman@cs.stanford.edu)
Newsgroups: sci.math
From: kekamins@zeno30.math.UWaterloo.ca (kekaminsky)
Subject: Re: Philosophy of Pi
Organization: University of Waterloo
Date: Sun, 13 Dec 1992 19:14:29 GMT

In article <1gd4dbINNm9q@gambier.rick.cs.ubc.ca> t4d192@rick.cs.ubc.ca (Bruce Paul Dow) writes:
>
>Of course, I have seen the Taylor series proof of Euler's
>identity but I can't seem to pull any intuitive feeling out 
>of it. It seems to me as well that having proved Euler's
>identity, you then have a strong link between e and Pi.
>Is this not so? I was wondering whether anyone knew how strong
>this link might be and whether we can express e and Pi in terms
>of each other in any useful way?
>

YESSSSS!!!!!  exp(i*Pi) + 1 = 0
is perhaps the most beautiful identity in mathematics, and it follows
directly from Euler's identity.

 The equation links the two most important transcendentals, the 'enigmatic' i
and of course of paramount importance 0 and 1, the basis of number.

As for 'intuitionalisms ' regarding the identity, think back to ordinary differential equations,
in particular, second order constant coefficient ones (ie. the nice easy ones).

Look at y''+y=0 and y''-y=0, as well as the mixed case, and think of the physical circumstances 
under which these arise-  SHM, RC circuits (or LC, or RL I can't remember).  Then look at the power
series soltns., or in particular the characteristic eqn. for y''+y=0 which as +-i as roots.

Hope this helps.
 
Kirk

Newsgroups: sci.math
From: solan@smauguio.no (Svein Olav G. Nyberg)
Subject: Re: Would Pi repeat, were it expressed in a base other than 10?
Reply-To: solan@smauguio.no (Svein Olav G. Nyberg)
Organization: University of Oslo, Norway
Date: Mon, 14 Dec 1992 16:04:21 GMT

In article <1992Dec14.152439.11715@news.eng.convex.com>, Dave Dodson
<dodson@convex.COM> writes:
|> In article <Bz4p1C.DtG@news.cso.uiuc.edu> brumleve@gauss (Dan
Brumleve) writes:
|> >bennett@math.ksu.edu (Andy Bennett) writes:
|> >
|> >>werdna@cco.caltech.edu (Andrew Tong) writes:
|> >>:Intuitively (not saying much, that is), I'd say yes, but that it
would re-
|> >>:quire a transcendental base....
|> >
|> >>Of course, pi to the base pi is 1.000000000....
|> >
|> >>:-)
|> >
|> >
|> >Not to nitpick, but it's 10.00000.... ;-)
|> 
|> Combining this with a previously-beaten-to-death result, we see that
|> pi to the base pi is 9.999... .

Base pi? No joking?

Well, what are the first 5 digits of what we normally
call "5" in this base?


Solan
From: tlea@ccwf.cc.utexas.edu (Thomas C. Lea)
Newsgroups: sci.math
Subject: Diff Eq Books
Date: 13 Dec 92 00:27:44 GMT
Organization: none
X-UserAgent: Nuntius v1.1.1d10
X-XXMessage-ID: <A74FE0153901B9B7@slip-x83.ots.utexas.edu>
X-XXDate: Sat, 12 Dec 92 02:29:41 GMT

I just finshed my first Diff Eq course and realy liked it.  To bad I had
such a horrible book, along with a teacher.  So I'm looking for a good
first book on Diff Eq.  The Book we used was by Guterman & Nitecki.  Any
suggestions would be greatly appreaciated.

				Thanks
From: arodgers@dcs.qmw.ac.uk (Angus H Rodgers)
Newsgroups: sci.math
Subject: A word problem
Date: 12 Dec 92 16:23:49 GMT
Organization: Computer Science Dept, QMW, University of London

Are there any finitely generated infinite semigroups in which
the idempotent law holds?

The "free idempotent semigroup" (if that's the name for it) on
2 generators has 6 elements. I don't know if the f.i.s. on 3
generators is infinite, but it's certainly quite big.

In trying to prove the infiniteness of the f.i.s. over a given
alphabet, say {a,b,c}, you can't build arbitrarily long reduced 
words by a haphazard process of pre- or post- multiplication by 
generators, because some reduced words do not occur as substrings
of any longer ones, e.g.:

                              abacaba
                          abacbabcabacbab
                       abcbabcabacabcbabcaba
                  abcbacabcbabcacbabcbacabcbabcac

Is there, nevertheless, a recipe or existence proof for arbitarily
long reduced words?
--
Gus Rodgers,  Dept. of Computer Science, | 
Queen Mary & Westfield College, Mile End | 
Road, London, England.   +44 71 975 5241 | 
E-mail (JANET):   arodgers@dcs.qmw.ac.uk | Post in haste, repent at leisure.
From: kosowsky@schottky.harvard.edu (Jeffrey J. Kosowsky)
Newsgroups: sci.math
Subject: Re: What is the number
Date: 17 Dec 92 17:50:53 GMT
Article-I.D.: schottky.KOSOWSKY.92Dec17125053
Organization: Harvard Robotics Lab, Harvard University

In article <BzE0v9.C1F@unccsun.uncc.edu> sbardhan@uncc.edu (Soumendu Bardhan) writes:

   Whose sum of the factorials of the digits is the number itself
   ( no programming please !).


Two examples come immediately to mind:  1
								145
Xref: sparky sci.math:16959 rec.puzzles:7933
From: columbus@strident.think.com (Michael Weiss)
Newsgroups: sci.math,rec.puzzles
Subject: The paradox of the second ace (was: Marilyn Vos Savant's error?)
Date: 15 Dec 92 11:12:55
Organization: Thinking Machines Corporation, Cambridge MA, USA
	<1992Dec15.012404.24027@galois.mit.edu>
	<1992Dec15.055832.26324@galois.mit.edu>
NNTP-Posting-Host: strident.think.com
In-reply-to: jbaez@riesz.mit.edu's message of Tue, 15 Dec 92 05:58:32 GMT

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.  Given that one
   is an ace, what's the probability that they are all aces?

   2)   You draw 4 cards from a well-shuffled standard deck.  Given that
   one is the ace of hearts, what's the probability that they are
   all aces?

   The answers to problems 1) and 2) are NOT THE SAME.

This is similar to the paradox of the second ace: a bridge player announces
he holds an ace.  What is the probability he holds another ace?  In a later
game, a bridge player announces he holds the ace of spades.  What is the
probability he holds another ace?  The second probability is greater than
the first.

(Assume randomly shuffled decks, and all that.)

Martin Gardner wrote into Marilyn Vos Savant with this paradox after the
famous "Monty Hall" brouhaha.

(Incidentally, John commented on how QM and probability are both
non-intuitive.  I'm too lazy to construct a bosonic version of the boy/girl
puzzle ("I've got two photons, one of them is a boy...") but it shouldn't
be too hard.)

Here is a proof of the inequality of the two probabilities that
avoids lots of grungy arithmetic:

Let:
  a = number of hands having exactly one ace 
  b = number of hands having two or more aces 
  s = number of hands having the ace of spades and no other aces
  t = number of hands having the ace of spades and at least one more ace

Then:
  prob(two or more aces | one ace) = b/(a+b)
  prob(two or more aces | ace of spades) = t/(s+t)

Now a=4s, as you can show by a simple combinatoric argument.  Also
b<2t, as you can show by a slightly more involved combinatoric
argument: let b = b2 + b3 + b4, where bi is the number of hands
having exactly i aces.  Also let t = t2 + t3 + t4, where ti is the
number of hands having exactly i aces, one of them the ace of spades.
Then

  b2 = 2 t2
  b3 = (4/3)t3
  b4 = t4

as simple combinatoric arguments will show.  So

  b =    b2 +      b3 + b4
    =   2t2 + (4/3)t3 + t4
    < 2( t2 +      t3 + t4) = 2t.

With a little algebra, and the weaker inequality b<4t, you can now show
that the first conditional probability is less than the second.

Incidentally, this argument does not use the fact that the deck contains 52
cards, or the fact that a bridge hand has 13 cards.  It does use the fact
that the deck contains 4 aces.  However, it can easily be generalized to
the case of a deck of cards with r suits, for any r >= 2.
Xref: sparky sci.math:17195 sci.physics:21447 sci.chem:5511
From: James_J_Kowalczyk@cup.portal.com
Newsgroups: sci.math,sci.physics,sci.chem
Subject: Re: TIME HAS INERTIA
Date: Fri, 18 Dec 92 17:30:38 PST
Organization: The Portal System (TM)
  <abian.724200923@pv343f.vincent.iastate.edu>
  <Bz7rtx.F77@knot.ccs.queensu.ca>

My, this topic is certainly getting a lot of bandwidth.

Can someone tell me who this Abian person is?
Why he seems to spell poorly?
Why he uses poor grammar?
Why he seems fascinated with CAPS and exclamation points?

His posts come across as so absurd as to be flame-bait or spoofs;
yet they generate a great deal of earnest discussion.

Thanks,
Jim Kowalczyk, Ph.D. 
Senior Research Chemist
Eisai Research Institute
Andover, MA  01810

James_J_Kowalczyk@cup.portal.com
Newsgroups: sci.math
From: kim@Lise.Unit.NO (Kim Gunnar St|vring \yhus)
Subject: Re: TIME HAS INERTIA
Organization: Norwegian Institute of Technology
Date: Sat, 19 Dec 92 22:20:13 GMT


Warning to new readers of Abians postings concerning his principle.

If you post something about his principle, no matter how polite you
formulate it, the usual thing for Abian is to either answer rudely and with
no real content, or ignore you.

It has been proved that his principle is informationless, unusable, and not
science, but he still lingers on.

Abian feeds on replies. Please don't feed him.

Kim0


Newsgroups: sci.math,sci.physics