HDTenEightyP/NetNews huggingface dataset:
From: zeno@athena.mit.edu (Richard Duffy)
Newsgroups: sci.math
Subject: equivalents of C.H. (was: Beloved Books + Request)
Date: 6 Sep 92 00:52:48 GMT
Organization: Massachvsetts Institvte of Technology

In article <ARA.92Sep2012009@camelot.ai.mit.edu>, Allan Adler writes:
>
> One of my favorite books is Sierpinski's Hypothese du continu, which studies
> the continuum hypothesis and its consequences and equivalents in analysis.

Ah, one of my favorites as well.  I believe it's even still in print from
Chelsea (N.Y.), though I forget whether it's the English translation that
they have.

Another of the results in that book is:  CH <==> there is a real function  f
such that the plane is a countable union of sets, each being either a
translate of the graph of  f  or a translate of  f's  graph rotated 90 degrees.
(Clearly we're talking about an extremely discontinuous function here).  I
would like to improve this result, if possible, to make the plane-covering
union be *disjoint*, but I have a notion this won't be possible ... any
guesses?

I can't refrain from bringing up again (though I haven't mentioned it here
in a long time) the following, which I happened across quite by chance once
when looking at another paper in some bound journal volume:

This is one of Paul Erdos's many delightful oddments: CH is equivalent to
the existence of an uncountable family  F  of entire functions (complex
functions holomorphic on the whole plane) such that for all points  z , the
set of images  {f(z) : f  in  F}  is countable.  Think about this in light of
the fact that any two distinct entire functions must take different values
at all but finitely many points in {1, 1/2, 1/3, 1/4, ...}, for example, yet
for each  1/n  the uncountably many  f's  take on only countably many values
f(1/n).  Pretty peculiar.     This equivalence is distinguished by the fact
that its proof (which is pretty short) uses bona-fide facts about convergent
power series---it's not just set theory in disguise.  It's not even
*real-variable* theory in disguise: there is an explicit uncountable family
of infinitely differentiable functions from  R  to  R  whose set of images at
each real number has cardinality at most 2! (an amusing exercise).



--
>> Richard Duffy  -----------\___ ((lambda (x) (list x (list (quote quote) x)))
Internet:  zeno@athena.mit.edu   \____  (quote (lambda (x)
Bitnet:    zeno%athena@MITVMA         \______(list x (list (quote quote) x)))))
Voicenet:  +1 617 253 4045                   \------------------ 
Xref: sparky sci.math:11398 comp.lang.misc:3065 comp.lang.fortran:3547 comp.std.c:2622 comp.lang.c:13633
From: hrubin@pop.stat.purdue.edu (Herman Rubin)
Newsgroups: sci.math,comp.lang.misc,comp.lang.fortran,comp.std.c,comp.lang.c
Subject: Re: Scientists as Programmers (was Re: Small Language Wanted)
Date: 14 Sep 92 13:09:52 GMT
Organization: Purdue University Statistics Department

In article <PRENER.92Sep14014510@prener.watson.ibm.com> prener@watson.ibm.com (Dan Prener) writes:

>In article <BuJ4Es.49w@mentor.cc.purdue.edu> hrubin@pop.stat.purdue.edu (Herman Rubin) writes:

>[... deletion ... (topic was obtaining long long product of two longs) ]

>>                                                      And it still is
>>not quite adequate.  One should be able to have the option of overloading
>>the * operator so that if c is long long, and a and b are long, this can
>>be written as c = a * b.  

>While that could be done, it goes very much against the spirit of most
>languages' expression evaluation, in which a * b has a value that is
>well defined without reference to the context in which it is used, in
>this case the type of the target of the assignment.

Now before the HLLs, the great bulk of computers were word oriented,
a word varying from 32-68 bits, and the general understanding was that
for integers a * b was some version of nxn -> 2n, and a/b was 2n/n
yielding quotient and remainder.  From that standpoint, it is the 
usual integer arithmetic which should be considered aberrant, and
the reduction to nxn -> n being a casting of the result.

Unfortunately, the hardware has been influenced by the languages, so
the logically natural computations now become a major programming
problem to do efficiently at all.
-- 
Herman Rubin, Dept. of Statistics, Purdue Univ., West Lafayette IN47907-1399
Phone: (317)494-6054
hrubin@pop.stat.purdue.edu (Internet, bitnet)  
{purdue,pur-ee}!pop.stat!hrubin(UUCP)
From: gjm11@cus.cam.ac.uk (G.J. McCaughan)
Newsgroups: sci.math
Subject: Re: A set-theoretic proof:need help
Date: 4 Sep 92 21:31:20 GMT
Organization: U of Cambridge, England

In article <181cvgINNptm@matt.ksu.ksu.edu>, bubai@matt.ksu.ksu.edu (P.Chatterjee)
writes:
> Hi,
> 
> I have a 'homework' problem which I'd like some help on. It's a proof. It's 
> awfully simple but the only problem is that I'm not sure how exactly one 
> should approach it especially if one is looking to give a 'slick' proof.
> 
> It goes like this:
> 
> Prove that for subsets A, B c X:
> 
> A c B <=> A U B = B <=> A ^ B = A <=> B' c A'
> 
> where: c:= subset
>        U:= union
>        ^:= intersection
> 
> My request is for some help on how to go about using the logical equivalence 
> symbol in proving the above equivalences. I have constructed a 'scrappy' and 
> lengthy proof but would like to achieve brevity in expression.
> 
> Would really appreciate if somebody, without giving the actual proof, helped me
> with how to proceed.
> 
> Thanks for all the help.

I'm not sure I can give much useful advice for this case without basically
giving you a complete proof. The trouble is that the most natural "proof" is
just to say "Well, it's obvious, isn't it?"...

One observation that is sometimes helpful (not sure whether it is in this case):
If you have to prove A<=>B<=>C...<=>Z, see if you can do it by proving something
like A=>B=>C=>...=>Z -- this can substantially decrease the number of separate
results you have to prove.

One other thing, specifically about this problem... (This will be really obvious
to some, but you mightn't have thought of things this way.) I'll put lots of
newlines in, in case it counts as telling you how to proceed.


































Things like union and intersection of sets are really sort of abstractions of
logical operations like "or" and "and". In other words, if A is the set of all
things such that P, and B is the set of all things such that Q, then AnB is
the set of all things such that (P and Q), and AuB is the set of all things 
such that (P or Q), and A c B if and only if P=>Q, and so on. Oh, and A=B
if and only if P<=>Q.

So the present problem is reduced to asking questions about one object at a
time: we have to prove P=>Q iff PvQ<=>Q iff etc.
Newsgroups: sci.math
From: gjm11@cus.cam.ac.uk (G.J. McCaughan)
Subject: Re: Permutation problem
Organization: U of Cambridge, England
Date: Fri, 4 Sep 1992 23:00:42 GMT

In article <Bu14Ep.IEE@matilda.vut.edu.au>, kcj@matilda.vut.edu.au (Kate) writes:

> Can someone tell me the formula for getting the number
> of permutations of any group of numbers that add up to a certain number. e.g.
> how many permutations of length 5 add up to 20?

Erm, what is a "permutation" in this context? Do you mean: how many strings
of numbers, no two of which are the same, add up to [whatever]?

If so...

These permutations come in "families" of size r! where r is the length of the
strings (because rearranging a permutation leaves it a permutation and doesn't
change what it adds up to). Dividing out by that, we want to know how many ways
there are to chop 20 up into 5 different parts, where we aren't bothered about
what order they come in.

This is the same as the number of ways to chop 10 = 20-(0+1+2+3+4) into 5 parts,
where they needn't be different and we don't care what order they come in. (We
can convert one type of "partition" into the other by subtracting 0 from the
smallest part, 1 from the next-smallest, and so on.)

I don't think there's a terribly nice general formula. If you write p(n,r)
for the number of ways to split n into r parts, where they don't have to be
all different, then ... um ... either the smallest part is 1 [p(n-1,r-1) of
these] or it's not and all the parts are at least 2 [p(n-r,r) of these].
So p(n,r) = p(n-1,r-1) + p(n-r,r), and you can use this (plus trivial things
about what happens when n or r is 0) to evaluate p(n,r) if n,r are small.
There's probably a rather more efficient way for large n...

Certainly p(n) [the sum of all the p(n,r)] is not very well understood. There
are recurrences, and estimates for large n, and a very few congruence relations;
and there *is* a formula that calculates it, but it's Very Strange and has
things like Bessel functions in it; but there's an awful lot that is not really
understood at all.
Newsgroups: sci.math
From: bryson@nas.nasa.gov (Steve T. Bryson)
Subject: Visualization 92 and VolViz92 call for student volunteers
Organization: Applied Research Office, NASA Ames Research Center
Date: 14 Sep 92 16:34:06
Distribution: sci.math



   Visualization '92 and the 1992 Boston Workshop on Volume Visualization
			
	****	CALL FOR STUDENT VOLUNTEERS		****

1992 Boston Workshop on Volume Visualization:	
		   Oct 17 and 18 1992, 	       Boston, Mass.
Visualization '92: 
		   Oct 18 through Oct 23 1992, Boston, Mass.

The Visualization '92 conference and the 1992 
Boston Workshop on Volume Visualization are seeking student volunteers. 
Volunteers will work  twenty hours or less in exchange for
complete free registration in the conference.  As the two
conferences are being held at the same location on
successive dates, the same students may volunteer  for both
conferences.  Come learn about the state of the art in 
volume visualization and the use of computer graphics to
study complex scientific phenomena.  

Student volunteers will be responsible for their own
transportation and lodging for the conferences.  To assist
in costs, limited shared rooms will be made available on a
first-come first-serve basis.  All students who are enrolled
full time in the fall 1992 are eligible.  

For more information, send dates of availability and 
field of study to:

For Visualization '92:
	Steve Bryson
	MS T045-1
	NASA Ames Research Center
	Moffett Field, Ca. 94035-1000
	(415) 604-4524 fax: (415) 604-3957
	bryson@nas.nasa.gov

For the Workshop on Volume Visualization:
	Roni Yagel
	Department of Computer and Information Science
	The Ohio State University
	228 Bolz Hall
	2036 Neil Av. Columbus, Ohio 43210-1277
	(614) 292-0060, fax: (614) 292-2911
	yagel@cis.ohio-state.edu




--

Steve Bryson
MS T045-1
NASA Ames Research Center
Moffett Field, Ca. 94035
(415) 604-4524
bryson@nas.nasa.gov
From: bakera@topaz.ucq.edu.au (Alan Baker)
Newsgroups: sci.math
Subject: Re: Help required Regression 
Date: 13 Sep 92 21:12:55 AET
Organization: University of Central Queensland, Australia

First of all let me appologise if this is the wrong news group to post 
this message. I am an Accountant by trade trying to treach myself some 
econometrics.

I have a problem I have not been able to solve and would certainly 
appreciate some help.

In the regression model 
y = X*B + e
by writing 
y = x*B1 + Xs*Bs + e
where x is a T-vector, B1 a scalar, Xs a T*(K-1) matrix and Bs is 
a (K-1) vector

Using partitioned matrix inversion, how is the result 

var(b1) = sigma^2 / summation(x-xbar)^2(1-R^2)    obtained

Where R^2 is the coefficient of determinantion obtained by regressing 
x on Xs

Any help at all would be much appreciated

================================================================================
Alan Baker                             AARNET:  BAKER@COVAX.COMMERCE.UQ.OZ.AU
Department of Commerce                 TELEPHONE: Work  07 365 6686
University of  Queensland, QLD 4072    International  +617 365 6686
Australia	                                   Home 07 371 1386
================================================================================
From: rpl@dsg.npl.co.uk (Richard Lampard)
Newsgroups: sci.math
Subject: Womble?
Date: 4 Sep 92 07:52:00 GMT
Organization: National Physical Laboratory, UK

Apologies in advance if this message sounds totally naive, but I am
not a mathemtician. Does anyone know anything about a technique
known as wombling? It is something to do with combining lots of
complex variables to obtain a single result, and was named after a
mathematician known as Womble.

Email replies to rpl@dsg.npl.co.uk would be appreciated.

Richard Lampard
National Physical Laboratory
Queens Road
Teddington
Middlesex
TW11 0LW
ENGLAND

Tel: +4481-943 7060
Fax: +4481-977 7091
Newsgroups: sci.math
From: tim@maths.tcd.ie (Timothy Murphy)
Subject: Re: Lebesgue integral (was: Couple of questions
Organization: Dept. of Maths, Trinity College, Dublin, Ireland.
Date: Sun, 13 Sep 1992 20:10:10 GMT

I find the implication that Riemann integration
is simpler than Lebesgue integration curious.

Lebesgue defined the integral as the area under the curve.
Since physicists and mathematicians all assume (at this level)
that "area" is a well-defined concept,
Lebesgue's definition is trivial, compared with Riemann's.
And if you want to be more rigorous,
you will have to define area anyway,
so the L-integral still retains its advantage.

The Riemann integral is of historic interest only.
Its continuing survival in mathematical education
is a monument to the conservatism of mathematicians.

-- 
Timothy Murphy  
e-mail: tim@maths.tcd.ie
tel: +353-1-2842366
s-mail: School of Mathematics, Trinity College, Dublin 2, Ireland
Newsgroups: sci.math
Subject: Proof of the impossibility of god.
From: wft@math.canterbury.ac.nz (Bill Taylor)
Date: 7 Sep 92 13:54:36 +1200
Distribution: world
Organization: Department of Mathematics, University of Canterbury

No, I haven't got one, but the endless spate of (I haven't read more than a
line or two of it, but probably) drivel on proofs of gods' existence, suggests
there should be a fair few of these ready to come forth.

And, after all, we all piously believe in democracy, civil liberties, and
EQUAL TIME for both sides, don't we ?
~~~~~~~~~~~~~~~~~~~
So; I'm sure there must be many posters who'd like to send in their own
"Proof of the impossibility of god".

As a starter, I'll throw out the idea that the set of numbers of halting machines can't be computable, yet, an omniscient god would know this set, and thus, 
(being rational (?) ), would have an algorithm to compute them, so they *would*
form a computable set......

Oh well.  I'm sure there are many arguments along logic/paradox lines that go
to prove the impossibility of god.     ;-)    Can't wait to see them !
-------------------------------------------------------------------------------
           Bill Taylor              wft@math.canterbury.ac.nz
-------------------------------------------------------------------------------
 Free will- the result of chaotic amplification of quantum events in the brain.
Galaxies- the result of chaotic amplification of quantum events in the big bang.  --------------------------------------------------------------------------------
From: shoham@ll.mit.edu (Daniel Shoham)
Newsgroups: sci.math
Subject: Re: Chess Problem
Date: 15 Sep 92 03:30:10 GMT
Organization: MIT Lincoln Laboratory

In article <1992Sep12.174545.23214@ima.isc.com> karl@ima.isc.com (Karl Heuer) writes:
>Here's a more advanced chessboard question.
>
>A white king, a black king, and a white pawn are randomly placed on an NxN
>chessboard.  (Independent, uniform distributions.)  What is the limit, as
>N -> infinity, of the probability that the position is a win for white?
>
>(Note that it doesn't matter what you do about collisions or illegal
>positions, since they occur with probability zero in the limit.)
>
>Karl Heuer   karl@ima.isc.com   uunet!ima!karl

We will assume optimal players on both sides (otherwise the question is
meaningless).

As stated, the probability is zero. A chess game is declared a draw
after 50 moves without an irreversible one. Even if white manages to get a
queen, the probability of finding the black king within 50 squares of an edge
is -> 0.

Suppose the 50-moves rule has been removed, and all that is needed is to queen
the pawn.

In a very large board, the likelihood of any of the three pieces starting
anywhere near any of the edges or each other is -> 0 so we will ignore it.

As I see it, white has a choice of two stratgies:
1. Run the pawn down and queen it without a king defense.
2. Move the king toward the pawn and defend it. Then advance the two in tandem.

(strategies where the king and pawn move toward each other may save time, but
they work just as well as strategy 2).

We will use an x-y system to describe the position of each square. The last
(queening) row is defined as y=0, whereas the first row is y=1. x, is similarly
ranged between 0 and 1.
(px,py) = pawn's starting position
(bx,by) = black (king) starting position
(wx,wy) = white (king) starting position

px, py, bx, by, wx, wy are iidrv with uniform (0,1) distribution.

Strategy 1 works if:
py < by  or  py < |px-bx|

Strategy 2 works if:
d(p,w) < d(p,b)      where d(i,j)=max(|ix-jx| , |iy-jy|)  (i.e. chess-distance)

We could now use brute force to integrate over all cases, or (if your religion
permits it) do a Monte Carlo simulation.

As my religion does not forbid me from doing a Monte Carlo after I have found
an analytic algorithm (that I am too lazy to integrate), I did it:

P(Draw) =~ 12.5%  (40,000 runs, i.e. +/- .5%)


Dan Shoham
shoham@ll.mit.edu
Newsgroups: sci.math
From: tycchow@riesz.mit.edu (Timothy Y. Chow)
Subject: Dominated Convergence Theorem
Organization: None.  This saves me from writing a disclaimer.
Date: Thu, 10 Sep 92 01:57:56 GMT

The DCT for real measurable functions states that if f_n -> f and there
exists integrable g such that |f_n| <= g for all n, then the limit of the
integrals equals the integral of the limit.

Here's a proof of DCT using Fatou's lemma.
1. Reduce to the nonnegative case by adding g to everything.
2. Applying Fatou to f_n implies integral(f) <= liminf integral(f_n).
3. Applying Fatou to g-f_n implies integral(f) >= limsup integral(f_n).

I haven't seen this proof in any book.  Why not?  It seems very simple and
intuitive.
-- 
Tim Chow     tycchow@math.mit.edu
Where a calculator on the ENIAC is equipped with 18,000 vacuum tubes and weighs
30 tons, computers in the future may have only 1,000 vacuum tubes and weigh
only 1 1/2 tons.                               ---Popular Mechanics, March 1949
Newsgroups: sci.math
From: rags@triples.math.mcgill.ca (Robert A. G. Seely)
Subject: Re: What's going on here ???
Organization: Dept Of Mathematics and Statistics
Date: Fri, 11 Sep 92 18:22:45 GMT


On this and other divergent series, see the end of Knopp's book Theory
and Application of Infinite Series.  
Newsgroups: sci.aeronautics
From: stote@tomcat.den.mmc.com (M. Stote)
Subject: Re: supersonic B-52's??
Organization: Martin Marietta Astronautics Group
Distribution: usa,local
Date: Sat, 12 Sep 1992 15:38:11 GMT


|> TW, I was at the (same) CNE airshow, and it was indeed a sight to
|> behold.  It did take quite a while to make its second pass, though.
|> I think by the time it finished its turn, it was over Buffalo :) !!
|> 
|> Gerald M
|> 
|> 
  I grew up near the UP (Upper Penninsula) of Michigan.  It is even more
impresive to have one roar over you on a low altitude penetration training
mission !

Mike Stoughton
Newsgroups: sci.aeronautics
From: jfb200@cbnewsd.cb.att.com (joseph.f.baugher)
Subject: Re: Ram Jet
Organization: AT&T Bell Laboratories -  Naperville, Illinois
Distribution: usa
Date: Wed, 16 Sep 1992 01:50:58 GMT
Summary: Ramjet-powered Shooting Star

In article <wb9omc.716504847@dynamo.ecn.purdue.edu>, wb9omc@dynamo.ecn.purdue.edu (Duane P Mantick) writes:
> 	I think they had some mixed film segments there.  The scenes
> where the ramjets were running I think were indeed all F80 stuff.

The  thread  on  the  ramjet-powered  P-80  Shooting  Star  piqued  my
curiosity  and  sent  me  off to my references.  I didn't find a whole
lot, but here's what I came up with.

There were two P-80A-1-LO fighters  that  were  experimentally  fitted
with  wingtip  ramjets.  44-85214  was  fitted  with a pair of 30-inch
diameter Marquardt C30-10B ramjets, whereas 44-85042 was fitted with a
pair  of 20-inch diameter Marquardt C20-85D ramjets.  The ramjets were
first used in flight on March 12, 1947, and on  June  17,  1948  P-80A
44-85214 flew briefly on ramjet power alone.

Another subject:  I seem to  remember a  French  experimental  ramjet-
powered aircraft which was launched from the top of a carrier aircraft.
I don't remember its name nor any  further details about it.   Perhaps
someone else remembers this plane.

Source:

  Lockheed Aircraft Since 1913,  Rene  J. Francillon, Naval  Institute
  Press, 1987.

Joe Baugher			**************************************
AT&T Bell Laboratories		*  "You see, something's going to    *
2000 North Naperville Road	*   happen.  Something wonderful!"   *
P. O. Box 3033			**************************************
Naperville, Illinois 60566-7033		
(708) 713 4548				
ihlpl!jfb                   Who, me?  Speak for AT&T?  Surely you jest!	
jfb200@cbnewsd.att.com
From: tpickett@auspex.com (Tom Pickett)
Newsgroups: sci.aeronautics
Subject: Re: F-23 Dead?
Date: 14 Sep 92 15:58:09 GMT
Reply-To: tpickett@auspex.com (Tom Pickett)
Organization: Auspex Systems, Inc.  Engineering

In article <CKD.92Sep13224759@loiosh.eff.org>, ckd@eff.org (Christopher 
Davis) writes:
> I didn't think the '15 had tip-mounts...

No it doesn't but I belive that this does some good on a/c that do have
wingtip mounts, such as the F-16...


Tom Pickett 
tpickett@auspex.com
SHO  GOZE
From: fiddler@concertina.Eng.Sun.COM (steve hix)
Newsgroups: sci.aeronautics
Subject: Re: F-23 Dead?
Date: 10 Sep 1992 18:42:58 GMT
Organization: Sun Microsystems, Mt. View, Ca.
NNTP-Posting-Host: concertina

In article <CKD.92Sep9170831@loiosh.eff.org> ckd@eff.org (Christopher Davis) writes:
>
>Also true, within reason.  Stuff like supercruise may or may not be
>useful; does anyone dogfight at Mach?  (*Can* anyone dogfight at Mach?)

Supercruise would be useful if for no other reason than that you can
get from base to combat area in less time with more usable fuel
remaining for fighting.  (You can also get faster turnaround for
successive missions.)  Assuming the pilots can handle more missions
per day, this means "more" aircraft available for use.

--
-----------------------------------------------------------------------
Of making many books there is no end...  |  What's wrong with chasing 
                       -- Eccl. 12:12    |  nebulae till 3 am?  |-(
-----------------------------------------------------------------------
From: SNOMCB@mvs.sas.com (Mike Bishop)
Newsgroups: sci.aeronautics
Subject: re: U2/TR1
Date: 12 Sep 92 21:26:00 GMT
Organization: SAS Institute Inc.

There is a very good book out that involves selection, training
and various tales of U2 pilots. The book is titled BLACK WATCH.
Can't remember the authors name off-hand, but I found it great
reading. 
 
 
  Mike